Question 14 Marks
Prove that: $\cos20^\circ\cos100^\circ+\cos100^\circ\cos140^\circ-\cos140^\circ\cos200^\circ=-\frac{3}{4}$
AnswerWe have, $\text{LHS}=\cos20^\circ\cos100^\circ+\cos100^\circ\cos140^\circ-\cos140^\circ\cos200^\circ$ $=\ \frac{1}{2}[2\cos100^\circ\cos20^\circ+2\cos140^\circ\cos100^\circ-2\cos200^\circ\cos140^\circ]$ $=\ \frac{1}{2}\big[\cos(100^\circ+20^\circ)+\cos(100^\circ-20^\circ)+\cos(140^\circ+100^\circ)\\ \ \ \ \ \ \ \ \ +\cos(140^\circ-100^\circ) -(\cos(200^\circ+140^\circ)+\cos(200^\circ-140^\circ))\big]$ $=\ \frac{1}{2}\big[\cos120^\circ+\cos80^\circ+\cos240^\circ+\cos40^\circ-\cos340^\circ-\cos60^\circ\big]$ $=\ \frac{1}{2}\Big[\cos(90^\circ+30^\circ)+\cos80^\circ+\cos40^\circ-\cos(180^\circ60^\circ)-\cos(360^\circ-20^\circ)-\frac{1}{2}\Big]$ $=\ \frac{1}{2}\Big[-\sin30^\circ+2\cos\Big(\frac{80^\circ+40^\circ}{2}\Big)\cos\Big(\frac{80^\circ-40^\circ}{2}\Big)-\cos60^\circ-\cos20^\circ-\frac{1}{2}\Big]$ $=\ \frac{1}{2}\Big[-\frac{1}{2}+2\cos60^\circ\cos20^\circ-\frac{1}{2}-\cos20^\circ-\frac{1}{2}$ $=\ \frac{1}{2}\Big[-\frac{3}{2}+2\times\frac{1}{2}\times\cos20^\circ-\cos20^\circ\Big]$ $=\ \frac{1}{2}\Big[-\frac{3}{2}+\cos20^\circ-\cos20^\circ\Big]$ $=\ \frac{1}{2}\Big[-\frac{3}{2}+0\Big]$ $=\ -\frac{3}{4}$ $=\ \text{RHS}$ $\therefore\ \cos20^\circ\cos100^\circ+\cos100^\circ\cos140^\circ-\cos140^\circ\cos200^\circ=-\frac{3}{4}.$ Hence proved.
View full question & answer→Question 24 Marks
Prove that: $\cos(\text{A+B+C})+\cos(\text{A}-\text{B+C})+\cos(\text{A+B}-\text{C})\\+\cos(-\text{A+B+C})=4\cos\text{A}\cos\text{B}\cos\text{C}$
AnswerWe have, $\text{LHS}=\cos(\text{A+B+C})+\cos(\text{A}-\text{B}+\text{C})\\ \ \ \ \ +\cos(\text{A+B}-\text{C})+\cos(-\text{A+B+C})$ $=\ [\cos(\text{A+B+C})+\cos(\text{A}-\text{B+C})]\\ \ \ \ \ +[\cos(\text{A+B}-\text{C})+\cos(-\text{A+B+C})$ $=\ 2\cos\Big\{\frac{\text{A+B+C+A}-\text{B}+\text{C}}{2}\Big\}\cos\Big\{\frac{\text{A+B+C}-\text{A+B}-\text{C}}{2}\Big\}\\ \ \ \ +2\begin{Bmatrix}\cos\Big\{\frac{\text{A+B}-\text{C}-\text{A+B+C}}{2}\Big\} \\\cos\Big\{\frac{\text{A+B}-\text{C}+\text{A}-\text{B}-\text{C}}{2}\Big\} \end{Bmatrix}$ $=\ 2\cos\Big(\frac{2\text{A}+2\text{C}}{2}\Big)\cos\Big(\frac{2\text{B}}{2}\Big)+2\cos\Big(\frac{2\text{B}}{2}\Big)\cos\Big\{\frac{2\text{A}-2\text{C}}{2}\Big\}$ $=\ 2\cos(\text{A+B})\cos(\text{B})+2\cos(\text{B})\cos(\text{A}-\text{C})$ $=\ 2\cos(\text{B})[\cos(\text{A+C})+\cos(\text{A}-\text{C})]$ $=\ 2\cos\text{B}\Big[2\cos\Big(\frac{\text{A+C+A}-\text{C}}{2}\Big)\cos\Big(\frac{\text{A+C}-\text{A+C}}{2}\Big)\Big]$ $=\ 2\cos(\text{B})[2\cos\text{A}\cos\text{C}]$ $=\ 4\cos\text{A}\cos\text{B}\cos{C}.$ $=\ \text{RHS}$
View full question & answer→Question 34 Marks
prove that: $\sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+\sin(\text{C}-\text{A})\cos(\text{B}-\text{D})+\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})=0$
AnswerWe have, $\text{LHS}=\sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+\sin(\text{C}-\text{A})\cos(\text{B}-\text{D})+\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})$ $=\ \frac{1}{2}[2\sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+\sin(\text{C}-\text{A})\cos(\text{B}-\text{D})+2\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})]$ $=\ \frac{1}{2}[\sin(\text{B}-\text{C}+\text{A}-\text{D})+\sin(\text{B}-\text{C}-\text{A+D})+\sin(\text{C}-\text{A}+\text{B}-\text{D})\\ \ \ \ \ +\sin(\text{C}-\text{A}-\text{B}+\text{D})+\sin(\text{A}-\text{B+C}-\text{D})+\sin(\text{A}-\text{B}-\text{C+D})]$ $=\ \frac{1}{2}[\sin(\text{A+B}-\text{C}-\text{D})+\sin(\text{B+D}-\text{C}-\text{A})+\sin(\text{B+C}-\text{A}-\text{D})\\ \ \ \ \ +\sin(\text{C+D}-\text{A}-\text{B})+\sin(\text{A+C}-\text{B}-\text{D})+\sin(\text{A+D}-\text{B}-\text{C})]$ $=\ \frac{1}{2}[\sin(\text{A+B}-\text{C}-\text{D})-\sin(\text{A+C}-\text{B}-\text{D})-\sin(\text{A+D}-\text{B}-\text{C})\\ \ \ \ \ -\sin(\text{A+B}-\text{C}-\text{D})+\sin(\text{A+C}-\text{B}-\text{D})+\sin(\text{A+D}-\text{B}-\text{C})]$ $=\ \frac{1}{2}[0]$ $=\ 0$ $=\ \text{RHS}$ $\sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+\sin(\text{C}-\text{A})\cos(\text{B}-\text{D})+\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})=0$ Hence proved.
View full question & answer→Question 44 Marks
$\text{If}\ \frac{\cos(\text{A}-\text{B})}{\cos(\text{A+B})}+\frac{\cos(\text{C+D})}{\cos(\text{C}-\text{D})}=0,$ prove that $\tan\text{A}\tan\text{B}\tan\text{C}\tan\text{D}=-1$
AnswerWe have, $\frac{\cos(\text{A}-\text{B})}{\cos(\text{A+B})}+\frac{\cos(\text{C+D})}{\cos(\text{C}-\text{D})}=0$ $\Rightarrow\ \frac{\cos(\text{A}-\text{B})}{\cos(\text{A+B})}=-\frac{\cos(\text{C+D})}{\cos(\text{C}-\text{D})}...(\text{i})$ Now, $\frac{\cos(\text{A}-\text{B})}{\cos(\text{A+B})}=-\frac{\cos(\text{C+D})}{\cos(\text{C}-\text{D})}$ $\Rightarrow\ \frac{\cos(\text{A}-\text{B})}{\cos(\text{A+B})}+1=\frac{-\cos(\text{C+D})}{\cos(\text{C}-\text{D})}+1$ $\Rightarrow\ \frac{\cos(\text{A}-\text{B})+\cos(\text{A+B})}{\cos(\text{A+B})}=\frac{-\cos(\text{C+D})+\cos(\text{C}-\text{D})}{\cos(\text{C}-\text{D})}$ $\Rightarrow\ \frac{\cos(\text{A}+\text{B})+\cos(\text{A}-\text{B})}{\cos(\text{A+B})}=\frac{-[\cos(\text{C+D})-\cos(\text{C}-\text{D})]}{\cos(\text{C}-\text{D})}...(\text{ii})$ Again, $\frac{\cos(\text{A}-\text{B})}{\cos(\text{A+B})}=-\frac{-\cos(\text{C+D})}{\cos(\text{C}-\text{D})}$ [By equation (i)] $\frac{\cos(\text{A}-\text{B})}{\cos(\text{A+B})}-1=-\frac{-\cos(\text{C+D})}{\cos(\text{C}-\text{D})}-1$ $\Rightarrow\ \frac{\cos(\text{A}-\text{B})-\cos(\text{A+B})}{\cos(\text{A+B})}=\frac{-\cos(\text{C+D})-\cos(\text{C}-\text{D})}{\cos(\text{C}-\text{D})}$ $\Rightarrow\ \frac{-(\cos(\text{A}+\text{B})-\cos(\text{A}-\text{B}))}{\cos(\text{A+B})}=\frac{-[\cos(\text{C+D})+\cos(\text{C}-\text{D})]}{\cos(\text{C}-\text{D})}$ $\Rightarrow\ \frac{\cos(\text{A}+\text{B})-\cos(\text{A}-\text{B}))}{\cos(\text{A+B})}=\frac{\cos(\text{C+D})+\cos(\text{C}-\text{D})}{\cos(\text{C}-\text{D})}...(\text{iii})$ Dividing equation (ii) by equation (iii), we get $\frac{\cos(\text{A}+\text{B})+\cos(\text{A}-\text{B})}{\cos(\text{A+B})-\cos(\text{A}-\text{B})}=-\frac{-[\cos(\text{C+D})-\cos(\text{C}-\text{D})]}{\cos(\text{C+D})+\cos(\text{C}-\text{D})}$ $\Rightarrow\ \frac{2\cos\Big\{\frac{\text{A+B+A}-\text{B}}{2}\Big\}\cos\Big\{\frac{\text{A+B}-\text{A+B}}{2}\Big\}}{-2\sin\Big\{\frac{\text{A+B+A}-\text{B}}{2}\Big\}\sin\Big\{\frac{\text{A+B}-\text{A+B}}{2}\Big\}}=\frac{-\Big[2\sin\Big\{\frac{\text{C+D+C}-\text{D}}{2}\Big\}\sin\Big\{\frac{\text{C+D}-\text{C+D}}{2}\Big\}\Big]}{2\cos\Big\{\frac{\text{C+D+C}-\text{D}}{2}\Big\}\cos\Big\{\frac{\text{C+D}-\text{C+D}}{2}\Big\}}$ $\Rightarrow\ \frac{\cos\text{A}\cos\text{B}}{-\sin\text{A}\sin\text{B}}=\frac{\sin\text{C}\sin\text{D}}{\cos\text{C}\cos\text{D}}$ $\Rightarrow\ \frac{1}{-\tan\text{A}\tan\text{B}}=\tan\text{C}\tan\text{D}$ $\Rightarrow\ -1=\tan\text{A}\tan\text{B}\tan\text{C}\tan\text{D}$ $\therefore\ \tan\text{A}\tan\text{B}\tan\text{C}\tan\text{D}=-1$ Hence proved.
View full question & answer→Question 54 Marks
$\text{If}\ \cos(\alpha+\beta)\sin(\gamma+\delta)=\cos(\alpha-\beta)\sin(\gamma-\delta),$ prove that $\cot\alpha\cot\beta\cot\gamma=\cot\delta$
AnswerWe have, $\cos(\alpha+\beta)\sin(\gamma+\delta)=\cos(\alpha-\beta)\sin(\gamma-\delta)$ $\Rightarrow\ \frac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=\frac{\sin(\gamma-\delta)}{\sin(\gamma+\delta)}...(\text{i})$ Now, $\frac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=\frac{\sin(\gamma-\delta)}{\sin(\gamma+\delta)}$ $\Rightarrow\ \frac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}+1=\frac{\sin(\gamma-\delta)}{\sin(\gamma+\delta)}+1$ $\Rightarrow\ \frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{\cos(\alpha-\beta)}=\frac{\sin(\gamma-\delta)+\sin(\gamma+\delta)}{\sin(\gamma+\delta)}...(\text{ii})$ Again, $\frac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}=\frac{\sin(\gamma-\delta)}{\sin(\gamma+\delta)}$ [By equation (i)] $\Rightarrow\ \frac{\cos(\alpha+\beta)}{\cos(\alpha-\beta)}-1=\frac{\sin(\gamma-\delta)}{\sin(\gamma+\delta)}-1$ $\Rightarrow\ \frac{\cos(\alpha+\beta)-\cos(\alpha-\beta)}{\cos(\alpha-\beta)}=\frac{\sin(\gamma-\delta)-\sin(\gamma+\delta)}{\sin(\gamma+\delta)}...(\text{iii})$ Dividing equation (ii) by equation (iii), we get $\frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{\cos(\alpha+\beta)-\cos(\alpha-\beta)}=\frac{\sin(\gamma-\delta)+\sin(\gamma+\delta)}{\sin(\gamma-\delta)-\sin(\gamma+\delta)}$ $\Rightarrow\ \frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{\cos(\alpha+\beta)-\cos(\alpha-\beta)}=-\Big[\frac{\sin(\gamma+\delta)+\sin(\gamma-\delta)}{\sin(\gamma+\delta)-\sin(\gamma-\delta)}\Big]$ $\Rightarrow\ \frac{2\cos\big\{\frac{\alpha+\beta+\alpha-\beta}{2}\big\}\cos\big\{\frac{\alpha+\beta-\alpha+\beta}{2}\big\}}{-2\sin\big\{\frac{\alpha+\beta+\alpha-\beta}{2}\big\}\sin\big\{\frac{\alpha+\beta-\alpha+\beta}{2}\big\}}\\ \ \ \ \ =-\Bigg[\frac{2\sin\big\{\frac{\gamma+\delta+\gamma-\delta}{2}\big\}\cos\big\{\frac{\gamma+\delta-\gamma+\delta}{2}\big\}}{2\sin\big\{\frac{\gamma+\delta-\gamma+\delta}{2}\big\}\cos\big\{\frac{\gamma+\delta+\gamma-\delta}{2}\big\}}\Bigg]$ $$$\Rightarrow\ \frac{\cos\alpha\cos\beta}{\sin\alpha\sin\beta}=\frac{\sin\gamma\cos\delta}{\sin\delta\cos\gamma}$ $\Rightarrow\ \cot\alpha\cot\beta=\frac{\sin\gamma\cos\delta}{\cos\gamma\sin\delta}$ $\Rightarrow\ \cot\alpha\cot\beta=\frac{\cot\delta}{\cot\gamma}$ $\Rightarrow\ \cot\alpha\cot\beta\cot\gamma=\cot\delta$ $\therefore\ \cot\alpha\cot\beta\cot\gamma=\cot\delta$ Hence proved.
View full question & answer→Question 64 Marks
Prove that: $\tan\text{x}\tan\Big(\frac{\pi}{3}-\text{x}\Big)\tan\Big(\frac{\pi}{3}+\text{x}\Big)=\tan3\text{x}$
Answer$\frac{\pi}{3}=60^\circ$ $\text{LHS}=\tan\text{x}\tan(60^\circ-\text{x})\tan(60^\circ+\text{x})$ $=\ \frac{\sin\text{x}\sin(60^\circ-\text{x})\sin(60^\circ+\text{x})}{\cos\text{x}\cos(60^\circ-\text{x})\cos(60^\circ+\text{x})}$ $=\ \frac{\sin\text{x}(\sin^260^\circ-\sin^2\text{x})}{\cos\text{x}(\cos^260^\circ-\sin^2\text{x})}$ $=\ \frac{\sin\text{x}\Big(\frac{3}{4}-\sin^2\text{x}\Big)}{\cos\text{x}\Big(\frac{1}{4}-\sin^2\text{x}\Big)}$ $=\ \frac{\sin\text{x}(3-4\sin^2\text{x})}{\cos\text{x}(1-4\sin^2\text{x})}$ $=\ \frac{3\sin\text{x}-4\sin^3\text{x}}{4\cos^3\text{x}-3\cos\text{x}}$ $=\ \frac{\sin3\text{x}}{\cos3\text{x}}$ $=\ \tan3\text{x}=\text{RHS}$
View full question & answer→Question 74 Marks
prove that: $\frac{\cos(\text{A+B+C})+\cos(-\text{A+B+C})+\cos(\text{A}-\text{B+C})+\cos(\text{A+B}-\text{C})}{\sin(\text{A+B+C})+\sin(-\text{A+B+C})+\sin(\text{A}-\text{B+C})-\sin(\text{A+B}-\text{C})}=\cot\text{C}$
AnswerWe have, $\text{LHS}=\frac{\cos(\text{A+B+C})+\cos(-\text{A+B+C})+\cos(\text{A}-\text{B+C})+\cos(\text{A+B}-\text{C})}{\sin(\text{A+B+C})+\sin(-\text{A+B+C})+\sin(\text{A}-\text{B+C})-\sin(\text{A+B}-\text{C})}$ $=\ \frac{2\cos\Big\{\frac{\text{A+B+C}-\text{A+B+C}}{2}\Big\}\cos\Big\{\frac{\text{A+B+C+A}-\text{B}-\text{C}}{2}\Big\}\\+2\cos\Big\{\frac{\text{A}-\text{B+C+A+B}-\text{C}}{2}\Big\}\cos\Big\{\frac{\text{A}-\text{B+C}-\text{A}-\text{B+C}}{2}\Big\}}{2\sin\Big\{\frac{\text{A+B+C}-\text{A+B+C}}{2}\Big\}\cos\Big\{\frac{\text{A+B+C+A}-\text{B}-\text{C}}{2}\Big\}\\+2\sin\Big\{\frac{\text{A}-\text{B+C}-\text{A}-\text{B+C}}{2}\Big\}\cos\Big\{\frac{\text{A}-\text{B+C+A+B}-\text{C}}{2}\Big\}}$ $=\ \frac{2\cos(\text{B+C})\cos\text{A}+2\cos\text{A}\cos(\text{C}-\text{B})}{2\sin(\text{B+C})\cos\text{A}+2\sin(\text{C}-\text{B})\cos\text{A}}$ $=\ \frac{2\cos\text{A}[\cos(\text{B+C})+\cos(\text{C}-\text{B})]}{2\cos\text{A}[\sin(\text{B+C})+\sin(\text{C}-\text{B})]}$ $=\ \frac{\cos(\text{B+C})+\cos(\text{C}-\text{B})}{\sin(\text{B+C})+\sin(\text{C}-\text{B})}$ $=\ \frac{2\cos\Big\{\frac{\text{B+C+C}-\text{B}}{2}\Big\}\cos\Big\{\frac{\text{B+C}-\text{C+B}}{2}\Big\}}{2\sin\Big\{\frac{\text{B+C+C}-\text{B}}{2}\Big\}\cos\Big\{\frac{\text{B+C}-\text{C+B}}{2}\Big\}}$ $=\ \frac{2\cos\text{C}\cos\text{B}}{2\sin\text{C}\cos\text{B}}$ $=\ \frac{\cos\text{C}}{\sin\text{C}}$ $=\ \cot\text{C}$ $=\ \text{RHS}$ $\therefore\ \frac{\cos(\text{A+B+C})+\cos(-\text{A+B+C})+\cos(\text{A}-\text{B+C})+\cos(\text{A+B}-\text{C})}{\sin(\text{A+B+C})+\sin(-\text{A+B+C})+\sin(\text{A}-\text{B+C})-\sin(\text{A+B}-\text{C})}=\cot\text{C}.$ Hence proved.
View full question & answer→Question 84 Marks
Prove that: $\sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)\\=4\sin\Big(\frac{\alpha+\beta}{2}\Big)\sin\Big(\frac{\beta+\gamma}{2}\Big)\sin\Big(\frac{\gamma+\alpha}{2}\Big)$
AnswerWe have, $\text{LHS}=\sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)$ $=\ (\sin\alpha+\sin\beta)+(\sin\gamma-\sin(\alpha+\beta+\gamma))$ $=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)+2\sin\Big(\frac{\gamma-(\alpha+\beta+\gamma)}{2}\Big)\cos\Big(\frac{\gamma+\alpha+\beta+\gamma}{2}\Big)$ $=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)+2\sin\Big(\frac{-\alpha-\beta}{2}\Big)\cos\Big(\frac{\alpha+\beta+2\gamma}{2}\Big)$ $=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)-2\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha+\beta+2\gamma}{2}\Big)$ $=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\Big[\cos\Big(\frac{\alpha-\beta}{2}\Big)-\cos\Big(\frac{\alpha+\beta+2\gamma}{2}\Big)\Big]$ $=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\Bigg[-2\sin\frac{\Big[\frac{\alpha-\beta}{2}+\frac{\alpha+\beta+2\gamma}{2}\Big]}{2}\sin\frac{\Big[\frac{\alpha-\beta}{2}-\frac{\alpha+\beta+2\gamma}{2}\Big]}{2}\Bigg]$ $$$=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\Big[-2\sin\Big[\frac{2\alpha+2\gamma}{2\times2}\Big]\sin\Big[\frac{-2\beta-2\gamma}{2\times2}\Big]\Big]$ $=\ -4\sin\Big(\frac{\alpha+\beta}{2}\Big)\Big[\sin\Big(\frac{\alpha+\gamma}{2}\Big)\sin\Big[\frac{-(\beta+\gamma)}{2}\Big]\Big]$ $=\ 4\sin\Big(\frac{\alpha+\beta}{2}\Big)\sin\Big(\frac{\alpha+\gamma}{2}\Big)\sin\Big(\frac{\beta+\gamma}{2}\Big)$ $=\ \text{RHS}$ $\sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)\\=4\sin\Big(\frac{\alpha+\beta}{2}\Big)\sin\Big(\frac{\beta+\gamma}{2}\Big)\sin\Big(\frac{\gamma+\alpha}{2}\Big)$ Hence proved.
View full question & answer→Question 94 Marks
Prove that: $\frac{\sin\text{A}\sin2\text{A}+\sin3\text{A}\sin6\text{A}}{\sin\text{A}\cos2\text{A}+\sin3\text{A}\cos6\text{A}}=\tan5\text{A}$
AnswerWe have, $\text{LHS}=\frac{\sin\text{A}\sin2\text{A}+\sin3\text{A}\sin6\text{A}}{\sin\text{A}\cos2\text{A}+\sin3\text{A}\cos6\text{A}}$ $=\ \frac{2[\sin\text{A}\sin2\text{A}+\sin3\text{A}\sin6\text{A}]}{2[\sin\text{A}\cos2\text{A}+\sin3\text{A}\cos6\text{A}]}$ $=\ \frac{2\sin2\text{A}\sin\text{A}+2\sin6\text{A}\sin3\text{A}}{2\cos2\text{A}\sin\text{A}+2\cos6\text{A}\sin3\text{A}}$ $=\ \frac{\cos(2\text{A}-\text{A})-\cos(2\text{A}+\text{A})+\cos(6\text{A}-3\text{A})-\cos(6\text{A}+3\text{A})}{\sin(2\text{A}+\text{A})-\sin(2\text{A}-\text{A})+\sin(6\text{A}+3\text{A})+\sin(6\text{A}-3\text{A})}$ $=\ \frac{\cos\text{A}-\cos3\text{A}+\cos3\text{A}-\cos9\text{A}}{\sin3\text{A}-\sin\text{A}+\sin9\text{A}-\sin3\text{A}}$ $=\ \frac{\cos\text{A}-\cos9\text{A}}{\sin9\text{A}-\sin\text{A}}$ $=\ \frac{-[\cos9\text{A}-\cos\text{A}]}{\sin9\text{A}-\sin\text{A}}$ $=\ \frac{-\Big(-2\sin\Big(\frac{9\text{A}+\text{A}}{2}\Big)\times\sin\Big(\frac{9\text{A}-\text{A}}{2}\Big)\Big)}{2\sin\Big(\frac{9\text{A}-\text{A}}{2}\Big)\times\cos\Big(\frac{9\text{A}+\text{A}}{2}\Big)}$ $=\ \frac{\sin5\text{A}\sin4\text{A}}{\sin4\text{A}\cos5\text{A}}$ $=\ \tan5\text{A}$ $=\ \text{RHS}$ $\therefore\ \frac{\sin\text{A}\sin2\text{A}+\sin3\text{A}\sin6\text{A}}{\sin\text{A}\cos2\text{A}+\sin3\text{A}\cos6\text{A}}=\tan5\text{A}$ Hence proved.
View full question & answer→Question 104 Marks
Show that:$\sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+\sin(\text{C}-\text{A})\\\cos(\text{B}-\text{D})+\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})=0$
AnswerWe have, $\text{LHS}=\ \sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+\sin(\text{C}-\text{A})\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \cos(\text{B}-\text{D})+\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})$ $=\ \frac{1}{2}[\sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+2\sin(\text{C}-\text{A})\\\ \ \ \ \ \ \ \ \cos(\text{B}-\text{D})+2\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})]$ $=\ \frac{1}{2}\Big[\sin(\text{B}-\text{C+A}-\text{D})+\sin(\text{B}-\text{C}-\text{A+D})+\sin(\text{C}-\text{A+B}-\text{D})\\\ \ \ \ \ \ \ +\sin(\text{C}-\text{A}-\text{B+D})+\sin(\text{A}-\text{B+C}-\text{D})-\sin(\text{A}-\text{B}-\text{C+D})\Big]$ $=\ \frac{1}{2}\Big[\sin(\text{A+B}-\text{C}-\text{D})+\sin(\text{B}-\text{C}-\text{A+D})+\sin(\text{C}-\text{A+B}-\text{D})\\\ \ \ \ \ \ +\sin(\text{C}-\text{A}-\text{B+D})+\sin(\text{A+C}-\text{B}-\text{D})+\sin(\text{A+D}-\text{B}-\text{C})\Big]$ $=\ \frac{1}{2}\Big[\sin(\text{A+B}-\text{C}-\text{D})+\sin(\text{B+D}-\text{C}-\text{A})+\sin(\text{-A+D}-\text{B}-\text{C})\\\ \ \ \ \ +\sin(-\text{A+B}-\text{C}-\text{D})+\sin(-\text{B+D}-\text{A}-\text{C})+\sin(\text{A+D}-\text{B}-\text{C})\Big]$ $=\ \frac{1}{2}\Big[\sin(\text{A+B}-\text{C}-\text{D})+\sin(\text{B+D}-\text{C}-\text{A})-\sin(\text{A+D}-\text{B}-\text{C})\\\ \ \ \ \ \ -\sin(\text{A+B}-\text{C}-\text{D})-\sin(\text{B+D}-\text{A}-\text{C})+\sin(\text{A+D}-\text{B}-\text{C})\Big]$ $=\ \frac{1}{2}\times0$$[\because\ \sin(-\theta)=-\sin\theta]$ $=\ 0$ $=\ \text{RHS}$ $\therefore\ \sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+\sin(\text{C}-\text{A})\\\ \ \ \ \ \cos(\text{B}-\text{D})+\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})=0$ Hence proved.
View full question & answer→Question 114 Marks
If $\alpha+\beta=\frac{\pi}{2},$show that the maximum value of $\cos\alpha\cos\beta\text{ is }\frac{1}{2}.$
Answer$\text{Let y}=\cos\alpha\cos\beta\text{ than},$ $\text{y}=\frac{1}{2}(2\cos\alpha\cos\beta)$ $=\ \frac{1}{2}[\cos(\alpha+\beta)+\cos(\alpha-\beta)]$ $=\ \frac{1}{2}[\cos90^\circ+\cos(\alpha-\beta)]$$[\because\ \alpha+\beta=90^\circ]$ $=\ \frac{1}{2}[0+\cos(\alpha-\beta)]$ $=\ \frac{1}{2}\cos(\alpha-\beta)$ $\Rightarrow\ \text{y}=\frac{1}{2}\cos(\alpha-\beta)$ Now, $-1\leq\cos(\alpha-\beta)\leq1$ $\Rightarrow\ \frac{-1}{2}\leq\frac{1}{2}\cos(\alpha-\beta)\leq\frac{1}{2}$ $\Rightarrow\ \frac{-1}{2}\leq\text{y}\leq\frac{1}{2}$ $\Rightarrow\ \frac{-1}{2}\leq\cos\alpha\cos\beta\leq\frac{1}{2}$ Hence, the maximum values of $\cos\alpha\cos\beta\text{ is }\frac{1}{2}. $
View full question & answer→Question 124 Marks
$\text{If}\ \cos(\text{A+B})\sin(\text{C}-\text{D})=\cos(\text{A}-\text{B})\sin(\text{C+D}),$ prove that $\tan\text{A}\tan\text{B}\tan\text{C}+\tan\text{D}=0$
AnswerWe have, $\cos(\text{A+B})\sin(\text{C}-\text{D})=\cos(\text{A}-\text{B})\sin(\text{C+D})$ $\Rightarrow\ \frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}...(\text{i})$ Now, $\frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}$ $\Rightarrow\ \frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}+1=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}+1$ $\Rightarrow\ \frac{\cos(\text{A+B})+\cos(\text{A}-\text{B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})+\sin(\text{C}-\text{D})}{\sin(\text{C}-\text{D})}...(\text{ii})$ Again, $\frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}$ [By equation (i)] $\Rightarrow\ \frac{\cos(\text{A+B})}{\cos(\text{A}-\text{B})}-1=\frac{\sin(\text{C+D})}{\sin(\text{C}-\text{D})}-1$ $\Rightarrow\ \frac{\cos(\text{A+B})-\cos(\text{A}-\text{B})}{\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})-\sin(\text{C}-\text{D})}{\sin(\text{C}-\text{D})}...(\text{iii})$ Dividing equation (ii) by equation (iii), we get $\frac{\cos(\text{A+B})+\cos(\text{A}-\text{B})}{\cos(\text{A+B})-\cos(\text{A}-\text{B})}=\frac{\sin(\text{C+D})+\sin(\text{C}-\text{D})}{\sin(\text{C+D})-\sin(\text{C}-\text{D})}$ $\Rightarrow\ \frac{2\cos\big\{\frac{\text{A+B+A}-\text{B}}{2}\big\}\cos\big\{\frac{\text{A+B}-\text{A+B}}{2}\big\}}{-2\sin\big\{\frac{\text{A+B+A}-\text{B}}{2}\big\}\sin\big\{\frac{\text{A+B}-\text{A+B}}{2}\big\}}\\ \ \ \ =\frac{2\sin\Big\{\frac{\text{C+D+C}-\text{D}}{2}\Big\}\cos\Big\{\frac{\text{C+D}-\text{C+D}}{2}\Big\}}{2\sin\Big\{\frac{\text{C+D}-\text{C+D}}{2}\Big\}\cos\Big\{\frac{\text{C+D+C}-\text{D}}{2}\Big\}}$ $\Rightarrow\ \frac{\cos\text{A}\cos\text{B}}{-\sin\text{A}\sin\text{B}}=\frac{\sin\text{C}\cos\text{D}}{\sin\text{D}\cos\text{C}}$ $\Rightarrow\ \frac{1}{-\tan\text{A}\tan\text{B}}=\frac{\sin\text{C}\cos\text{D}}{\cos\text{C}\sin\text{D}}$ $\Rightarrow\ \frac{-1}{\tan\text{A}\tan\text{B}}=\frac{\tan\text{C}}{\tan\text{D}}$ $\Rightarrow\ -\tan\text{D}=\tan\text{A}\tan\text{B}\tan\text{C}$ $\Rightarrow\ \tan\text{A}\tan\text{B}\tan\text{C}=-\tan\text{D}$ $\Rightarrow\ \tan\text{A}\tan\text{B}\tan\text{C}+\tan\text{D}=0$ Hence proved.
View full question & answer→Question 134 Marks
Prove that: $\sin\text{A}+\sin2\text{A}+\sin4\text{A}+\sin5\text{A}=4\cos\frac{\text{A}}{2}\cos\frac{2\text{A}}{2}\cos4\text{A}$
AnswerWe have, $\text{LHS}=\sin\text{A}+\sin2\text{A}+\sin4\text{A}+\sin5\text{A}$ $=\ (\sin2\text{A}+\sin\text{A})+(\sin5\text{A}+\sin4\text{A})$ $=\ \Big[2\sin\Big(\frac{2\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{2\text{A}-\text{A}}{2}\Big)\Big]+\Big[2\sin\Big(\frac{5\text{A}+4\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}-4\text{A}}{2}\Big)\Big]$ $=\ 2\sin\frac{3\text{A}}{2}\cos\frac{\text{A}}{2}+2\sin\frac{9\text{A}}{2}\cos\frac{\text{A}}{2}$ $=\ 2\cos\frac{\text{A}}{2}\Big[\sin\frac{3\text{A}}{2}+\sin\frac{9\text{A}}{2}\Big]$ $=\ 2\cos\frac{\text{A}}{2}\Big[\sin\frac{9\text{A}}{2}+\sin\frac{3\text{A}}{2}\Big]$ $=\ 2\cos\frac{\text{A}}{2}\Big[2\sin\Big\{\frac{1}{2}\Big(\frac{9\text{A}}{2}+\frac{3\text{A}}{2}\Big)\Big\}\cos\Big\{\frac{1}{2}\Big(\frac{9\text{A}}{2}-\frac{3\text{A}}{2}\Big)\Big\}\Big]$ $=\ 4\cos\frac{\text{A}}{2}\Big[\sin\frac{12\text{A}}{4}\cos\frac{6\text{A}}{4}\Big]$ $=\ 4\cos\frac{\text{A}}{2}\sin3\text{A}\cos\frac{3\text{A}}{2}$ $=\ 4\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}\sin3\text{A}$ $=\ \text{RHS}$ $\therefore\ \sin\text{A}+\sin2\text{A}+\sin4\text{A}+\sin5\text{A}=4\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}\sin3\text{A}.$ Hence proved.
View full question & answer→Question 144 Marks
Prove that: $\tan20^\circ\tan40^\circ\tan60^\circ\tan80^\circ=3$
Answer$\text{LHS}=\tan20^\circ\tan40^\circ\tan60^\circ\tan80^\circ$ $=\ (\tan20^\circ\tan40^\circ\tan80^\circ)\sqrt3$$[\because\ \tan60^\circ=\sqrt3]$ $=\ \Big(\frac{\sin20^\circ\sin40^\circ\sin80^\circ}{\cos20^\circ\cos40^\circ\cos80^\circ}\Big)\sqrt3$ $=\ \frac{(2\sin20^\circ\sin40^\circ)\sin80^\circ\times\sqrt3}{(2\cos20^\circ\cos40^\circ)\cos80^\circ}$ Applying $\Rightarrow\ 2\sin\text{A}\sin\text{B}-\cos(\text{A}-\text{B})-\cos(\text{A+B})$ $2\cos\text{A}\cos\text{B}-\cos(\text{A+B})+\cos(\text{A}-\text{B})$ $=\ \frac{(\cos(40^\circ-20^\circ)-\cos(40^\circ+20^\circ))\sin80^\circ\times\sqrt3}{(\cos(20^\circ+40^\circ)+\cos(40^\circ-20^\circ))\cos80^\circ}$ $=\ \frac{(\cos20^\circ-\cos60^\circ)\sin80^\circ\times\sqrt3}{(\cos60^\circ+\cos20^\circ)\cos80^\circ}$ $=\ \frac{\Big(\cos20^\circ-\frac{1}{2}\Big)\sin80^\circ\times\sqrt3}{\Big(\frac{1}{2}+\cos20^\circ\Big)\cos80^\circ}$ $=\ \frac{(2\sin80^\circ\cos20^\circ-\sin80^\circ)\sqrt3}{\cos80^\circ+2\cos20^\circ\cos80^\circ}$ $\Rightarrow\ 2\sin\text{A}\cos\text{B}-\sin(\text{A+B})+\sin(\text{A}-\text{B})$ $=\ \frac{(\sin(80^\circ+20^\circ)+\sin(80^\circ-20^\circ)-\sin80^\circ)\sqrt3}{\cos80^\circ+(\cos(20^\circ+80^\circ)+\cos(80^\circ-20^\circ))}$ $=\ \frac{(\sin100^\circ+\sin60^\circ-\sin80^\circ)\sqrt3}{\cos80^\circ+\cos(180^\circ-80^\circ)+\cos60^\circ}$ $=\ \frac{\Big(\sin(180^\circ+\frac{\sqrt3}{2}-\sin80^\circ)\Big)\sqrt3}{\cos80^\circ-\cos80^\circ+\cos60^\circ}$ $=\ \frac{\frac{3}{2}}{\frac{1}{2}}=3=\text{RHS}$
View full question & answer→Question 154 Marks
Prove that: $\sin10^\circ\sin50^\circ\sin60^\circ\sin70^\circ=\frac{\sqrt3}{16}$
Answer$\text{LHS}=\sin10^\circ\sin50^\circ\sin60^\circ\sin70^\circ$ $\Big(\sin10^\circ\sin50^\circ\sin70^\circ\frac{\sqrt3}{2}\Big)$$\Big[\because\ \sin60^\circ=\frac{\sqrt3}{2}\Big]$ $=\ \sin(90^\circ-80^\circ)\sin(90^\circ-40^\circ)\sin(90^\circ-20^\circ)\frac{\sqrt3}{2}$ $=\ \cos80^\circ\cos40^\circ\cos20^\circ\frac{\sqrt3}{2}$ $=\ \frac{\sqrt3}{2\times2}(2\cos40^\circ\cos20^\circ)\cos80^\circ$$[\because\ 2\cos\text{A}\cos\text{B}=(\cos\text{A+B})+\cos(\text{A}-\text{B})]$ $=\ \frac{\sqrt3}{2\times2}[\cos(40^\circ+20^\circ)+\cos(40^\circ-20^\circ)]\cos80^\circ$ $=\ \frac{\sqrt3}{2\times2}[\cos60^\circ+\cos20^\circ]\cos80^\circ$ $=\ \frac{\sqrt3}{2\times2}\Big[\frac{1}{2}+\cos20^\circ\Big]\cos80^\circ$ $=\ \frac{\sqrt3}{4}\Big[\frac{1}{2}\cos80^\circ+\cos20^\circ\cos80^\circ\Big]$ $=\ \frac{\sqrt3}{8}[\cos80^\circ+2\cos20^\circ\cos80^\circ]$ $=\ \frac{\sqrt3}{8}[\cos80^\circ+\cos(80^\circ+20^\circ)+\cos(80^\circ-20^\circ)]$ $=\ \frac{\sqrt3}{8}[\cos80^\circ+\cos100^\circ+\cos60^\circ]$ $=\ \frac{\sqrt3}{8}[\cos80^\circ+\cos(180^\circ-80^\circ)+\cos60^\circ]$ $=\ \frac{\sqrt3}{8}[\cos60^\circ]=\frac{\sqrt3}{16}=\text { RHS}$
View full question & answer→Question 164 Marks
Prove that: $\frac{\sin3\text{A}+\sin5\text{A}+\sin7\text{A}+\sin9\text{A}}{\cos3\text{A}+\cos5\text{A}+\cos7\text{A}\cos9\text{A}}=\tan6\text{A}$
AnswerWe have, $\text{LHS}=\frac{\sin3\text{A}+\sin5\text{A}+\sin7\text{A}+\sin9\text{A}}{\cos3\text{A}+\cos5\text{A}+\cos7\text{A}+\cos9\text{A}}$ $=\ \frac{(\sin9\text{A}+\sin3\text{A})+(\sin7\text{A}+\sin5\text{A})}{(\cos9\text{A}+\cos3\text{A})+(\cos7\text{A}+\cos5\text{A})}$ $=\ \frac{2\sin\Big(\frac{9\text{A}+3\text{A}}{2}\Big)\cos\Big(\frac{9\text{A}-3\text{A}}{2}\Big)+2\sin\Big(\frac{9\text{A}-3\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}-5\text{A}}{2}\Big)}{2\cos\Big(\frac{9\text{A}+3\text{A}}{2}\Big)\cos\Big(\frac{9\text{A}-3\text{A}}{2}\Big)+2\cos\Big(\frac{7\text{A}+5\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}-5\text{A}}{2}\Big)}$ $=\ \frac{2\sin6\text{A}\cos3\text{A}+2\sin6\text{A}\cos\text{A}}{2\cos6\text{A}\cos3\text{A}+2\cos6\text{A}\cos\text{A}}$ $=\ \frac{2\sin6\text{A}(\cos3\text{A}+\cos\text{A})}{2\cos6\text{A}(\cos3\text{A}+\cos\text{A})}$ $=\ \frac{\sin6\text{A}}{\cos6\text{A}}$ $=\ \tan6\text{A}$ $=\ \text{RHS}$ $\therefore\ \frac{\sin3\text{A}+\sin5\text{A}+\sin7\text{A}+\sin9\text{A}}{\cos3\text{A}+\cos5\text{A}+\cos7\text{A}\cos9\text{A}}=\tan6\text{A}$ Hence proved.
View full question & answer→Question 174 Marks
Prove that: $\frac{\sin5\text{A}\cos2\text{A}-\sin6\text{A}\cos\text{A}}{\sin\text{A}\sin2\text{A}-\cos2\text{A}\cos3\text{A}}=\tan\text{A}$
AnswerWe have, $\text{LHS}=\frac{\sin5\text{A}\cos2\text{A}-\sin6\text{A}\cos\text{A}}{\sin\text{A}\sin2\text{A}-\cos2\text{A}\cos3\text{A}}$ $=\ \frac{2(\sin5\text{A}\cos2\text{A}-\sin6\text{A}\cos\text{A})}{2(\sin\text{A}\sin2\text{A}-\cos2\text{A}\cos3\text{A})}$ $=\ \frac{2\sin5\text{A}\cos2\text{A}-2\sin6\text{A}\cos\text{A}}{2\sin\text{A}\sin2\text{A}-2\cos2\text{A}\cos3\text{A}}$ $=\ \frac{\sin(5\text{A}+2\text{A})+\sin(5\text{A}-2\text{A})-[\sin(6\text{A}+\text{A})+\sin(6\text{A}-\text{A})]}{\cos(2\text{A}-\text{A})-\cos(2\text{A}+\text{A})-[\cos(3\text{A}+2\text{A})+\cos(3\text{A}-2\text{A})]}$ $=\ \frac{\sin7\text{A}+\sin3\text{A}-\sin7\text{A}-\sin5\text{A}}{\cos\text{A}-\cos3\text{A}-\cos5\text{A}-\cos\text{A}}$ $=\ \frac{\sin3\text{A}-\sin5\text{A}}{-\cos3\text{A}-\cos5\text{A}}$ $=\ \frac{-(\sin5\text{A}-\sin3\text{A})}{-(\cos5\text{A}+\cos3\text{A})}$ $=\ \frac{\sin5\text{A}-\sin3\text{A}}{\cos5\text{A}+\cos3\text{A}}$ $=\ \frac{2\sin\Big(\frac{5\text{A}-3\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}+3\text{A}}{2}\Big)}{2\cos\Big(\frac{5\text{A}+3\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}-3\text{A}}{2}\Big)}$ $=\ \frac{\sin\text{A}\cos4\text{A}}{\cos4\text{A}\cos\text{A}}$ $=\ \frac{\sin\text{A}}{\cos\text{A}}$ $=\ \tan\text{A}$ $=\ \text{RHS}$ $\therefore\ \frac{\sin5\text{A}\cos2\text{A}-\sin6\text{A}\cos\text{A}}{\sin\text{A}\sin2\text{A}-\cos2\text{A}\cos3\text{A}}=\tan\text{A}$ Hence proved.
View full question & answer→Question 184 Marks
Prove that: $\cos20^\circ\cos40^\circ\cos80^\circ=\frac{1}{8}$
Answer$\text{LHS}=\cos20^\circ\cos40^\circ\cos80^\circ$ $=\ \frac{1}{2}(2\cos20^\circ\cos40^\circ)\cos80^\circ$ $=\ \frac{1}{2}[\cos(40^\circ+20^\circ)+\cos(40^\circ-20^\circ)]\cos80^\circ$$[\because\ 2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})]$ $=\ \frac{1}{2}[\cos60^\circ+\cos20^\circ]\cos80^\circ$ $=\ \frac{1}{2}\Big[\frac{1}{2}+\cos20^\circ\Big]\cos80^\circ$ $=\ \frac{1}{2}[\cos80^\circ+2\cos20^\circ\cos80^\circ]$ $=\ \frac{1}{4}[\cos80^\circ+\cos(80^\circ+20^\circ)+\cos(20^\circ-80^\circ)]$ $=\ \frac{1}{4}[\cos80^\circ+\cos100^\circ+\cos60^\circ]$ $=\ \frac{1}{4}[\cos80^\circ+\cos(180^\circ-80^\circ)+\cos60^\circ]$ $=\ \frac{1}{4}[\cos80^\circ-\cos80^\circ+\cos60^\circ]$ $= \frac{1}{4}\Big[\frac{1}{2}\Big]=\frac{1}{8}=\text{RHS}$
View full question & answer→Question 194 Marks
Prove that: $4\cos\text{x}\cos\Big(\frac{\pi}{3}+\text{x}\Big)\cos\Big({\frac{\pi}{3}-\text{x}}\Big)=\cos3\text{x}$
AnswerWe have, $\text{LHS}=4\cos\text{x}\cos\Big(\frac{\pi}{3}+\text{x}\Big)\cos\Big(\frac{\pi}{3}-\text{x}\Big)$ $=\ 2\cos\text{x}\Big[2\cos\Big(\frac{\pi}{3}+\text{x}\Big)\cos\Big(\frac{\pi}{3}-\text{x}\Big)\Big]$ $=\ 2\cos\text{x}\Big[2\cos\Big(\frac{\pi}{3}+\text{x}+\frac{\pi}{3}-\text{x}\Big)+\cos\Big(\frac{\pi}{3}+\text{x}-\frac{\pi}{3}+\text{x}\Big)\Big]$ $=\ 2\cos\text{x}\Big[\cos\frac{2\pi}{3}+\cos2\text{x}\Big]$ $=\ 2\cos\text{x}\Big[\cos\Big(\frac{\pi}{2}+\frac{\pi}{6}\Big)+\cos2\text{x}\Big]$ $=\ 2\cos\text{x}\Big[-\sin\frac{\pi}{6}+\cos2\text{x}\Big]$ $=\ 2\cos\text{x}\Big[-\frac{1}{2}+\cos2\text{x}\Big]$ $=\ -2\cos\text{x}\times\frac{1}{2}+2\cos\text{x}\cos2\text{x}$ $=\ -\cos\text{x}+[\cos(\text{x}+2\text{x})+\cos(2\text{x}-\text{x})]$ $=\ -\cos\text{x}+\cos3\text{x}+\cos\text{x}$ $=\ \cos3\text{x}$ $=\ \text{RHS}$ LHS = RHS Hence Proved.
View full question & answer→Question 204 Marks
Prove that: $\cos\text{A}+\cos3\text{A}+\cos5\text{A}+\cos7\text{A}=4\cos\text{A}\cos2\text{A}\cos4\text{A}$
AnswerWe have, $\text{LHS}=\cos\text{A}+\cos3\text{A}+\cos5\text{A}+\cos7\text{A}$ $=\ [\cos3\text{A}+\cos\text{A}]+[\cos7\text{A}+\cos5\text{A}]$ $=\ \Big[2\cos\Big(\frac{3\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{3\text{A}-\text{A}}{2}\Big)\Big]+\Big[2\cos\Big(\frac{7\text{A}+5\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}-5\text{A}}{2}\Big)\Big]$ $=\ 2\cos2\text{A}\cos\text{A}+2\cos6\text{A}\cos4\text{A}$ $=\ 2\cos\text{A}[\cos2\text{A}+\cos6\text{A}]$ $=\ 2\cos\text{A}[\cos2\text{A}+\cos6\text{A}]$ $=\ 2\cos2\text{A}\Big[2\cos\Big(\frac{6\text{A}+2\text{A}}{2}\Big)\cos\Big(\frac{6\text{A}-\text{A}}{2}\Big)\Big]$ $=\ 4\cos\text{A}[\cos4\text{A}\cos2\text{A}]$ $=\ \text{RHS}$ $\therefore\ \cos\text{A}+\cos3\text{A}+\cos5\text{A}+\cos7\text{A}=4\cos\text{A}\cos2\text{A}\cos4\text{A}$ Hence proved.
View full question & answer→Question 214 Marks
Prove that: $\frac{\sin11\text{A}\sin\text{A}+\sin7\text{A}\sin3\text{A}}{\cos11\text{A}\sin\text{A}+\cos7\text{A}\sin3\text{A}}=\tan8\text{A}$
AnswerWe have, $\text{LHS}=\frac{\sin11\text{A}\sin\text{A}+\sin7\text{A}\sin3\text{A}}{\cos11\text{A}\sin\text{A}+\cos7\text{A}\sin3\text{A}}$ $=\ \frac{2(\sin11\text{A}\sin\text{A}+\sin7\text{A}\sin3\text{A})}{2(\cos11\text{A}\sin\text{A}+\cos7\text{A}\sin3\text{A})}$ $=\ \frac{2\sin11\text{A}\sin\text{A}+2\sin7\text{A}\sin3\text{A}}{2\sin11\text{A}\sin\text{A}+2\cos7\text{A}\sin3\text{A}}$ $=\ \frac{\cos(11\text{A}-\text{A})-\cos(11\text{A}+\text{A})+\cos(7\text{A}-3\text{A})-\cos(7\text{A}+3\text{A})}{\sin(11\text{A}+\text{A})-\sin(11\text{A}-\text{A})+\sin(7\text{A}+3\text{A})-\sin(7\text{A}-3\text{A})]}$ $=\ \frac{\cos10\text{A}-\cos12\text{A}+\cos4\text{A}-\cos10\text{A}}{\sin12\text{A}-\sin10\text{A}+\sin10\text{A}-\sin4\text{A}}$ $=\ \frac{-(\cos12\text{A}-\cos4\text{A})}{\sin12\text{A}-\sin4\text{A}}$ $=\ \frac{-\Big[2\sin\Big(\frac{12\text{A}+4\text{A}}{2}\Big)\sin\Big(\frac{12\text{A}-4\text{A}}{2}\Big)\Big]}{2\sin\Big(\frac{12\text{A}-4\text{A}}{2}\Big)\cos\Big(\frac{12\text{A}+4\text{A}}{2}\Big)}$ $=\ \frac{2\sin8\text{A}\sin4\text{A}}{2\sin4\text{A}\cos8\text{A}}$ $=\ \frac{\sin8\text{A}}{\cos8\text{A}}$ $=\ \tan8\text{A}$ $=\ \text{RHS}$ $\therefore\ \frac{\sin11\text{A}\sin\text{A}+\sin7\text{A}\sin3\text{A}}{\cos11\text{A}\sin\text{A}+\cos7\text{A}\sin3\text{A}}=\tan8\text{A}$ Hence proved.
View full question & answer→Question 224 Marks
Prove that: $\frac{\sin\text{A}+\sin\text{B}}{\sin\text{A}-\sin\text{B}}=\tan\Big(\frac{\text{A}-\text{B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)$
AnswerWe have, $\text{LHS}=\frac{\sin\text{A}+\sin\text{B}}{\sin\text{A}-\sin\text{B}}$ $=\ \frac{2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}{2\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)}$ $=\ \frac{\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}{\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)}$ $= \tan\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)$ $=\ \text{RHS}$ $\therefore\ \frac{\sin\text{A}+\sin\text{B}}{\sin\text{A}-\sin\text{B}}=\tan\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big).$ Hence proved.
View full question & answer→Question 234 Marks
Prove that: $\frac{\cos\text{A}+\cos\text{B}}{\cos\text{B}-\cos\text{A}}=\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)$
AnswerWe have, $\text{LHS}=\frac{\cos\text{A}+\cos\text{B}}{\cos\text{B}-\cos\text{A}}$ $=\ \frac{2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}{-2\sin\Big(\frac{\text{B}+\text{A}}{2}\Big)\sin\Big(\frac{\text{B}-\text{A}}{2}\Big)}$ $=\ \frac{-\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}{\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{B}-\text{A}}{2}\Big)}$ $=\ \frac{-\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}{-\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\sin\Big(\frac{\text{A}-\text{B}}{2}\Big)}$ $= \cot\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big)$ $=\ \text{RHS}$ $\therefore\ \frac{\cos\text{A}+\cos\text{B}}{\cos\text{B}-\cos\text{A}}=\cot\Big(\frac{\text{A+B}}{2}\Big)\cot\Big(\frac{\text{A}-\text{B}}{2}\Big).$ Hence proved.
View full question & answer→Question 244 Marks
Show that: $\sin25^\circ\cos115^\circ=\frac{1}{2}(\sin140^\circ-1)$
Answer$\text{LHS}=\ \sin25^\circ\cos115^\circ$ $=\ \frac{2\sin25^\circ\cos115^\circ}{2}$ We Know that $2\sin\text{A}\cos\text{B}=\sin(\text{A+B})+\sin(\text{A}-\text{B})$ $=\ \frac{1}{2}[\sin(25^\circ+115^\circ)+\sin(25^\circ-115^\circ)]$ $=\ \frac{1}{2}[\sin140^\circ+\sin(-90^\circ)]$ $\sin(-\theta)=-\sin\theta$ $\text{And},\sin(90^\circ+\theta)=\cos\theta$ $\Rightarrow\ \frac{1}{2}[\sin(90^\circ+50^\circ)-\sin90^\circ]$ $=\ \frac{1}{2}[\cos50^\circ-1]$ Also, $\cos\theta=\sin(90^\circ+\theta)$ $\cos50^\circ=\sin(90^\circ+50^\circ)=\sin140^\circ$ $\frac{1}{2}[\sin140^\circ-1]=\text{RHS}$
View full question & answer→Question 254 Marks
Prove that: $\frac{\sin(\theta+\phi)-2\sin\theta+\sin(\theta-\phi)}{\cos(\theta+\phi)-2\cos\theta+\cos(\theta-\phi)}=\tan\theta$
AnswerWe have, $\text{LHS}=\frac{\sin(\theta+\phi)-2\sin\theta+\sin(\theta-\phi)}{\cos(\theta+\phi)-2\cos\theta+\cos(\theta-\phi)}$ $=\ \frac{\sin(\theta+\phi)+\sin(\theta-\phi)-2\sin\theta}{\cos(\theta+\phi)+\cos(\theta-\phi)-2\cos\theta}$ $=\ \frac{2\sin\Big[\frac{(\theta+\phi)+(\theta-\phi)}{2}\Big]\cos\Big[\frac{(\theta+\phi)-(\theta-\phi)}{2}\Big]-2\sin\theta}{2\cos\Big[\frac{(\theta+\phi)+(\theta-\phi)}{2}\Big]\cos\Big[\frac{(\theta+\phi)-(\theta-\phi)}{2}\Big]-2\cos\theta}$ $=\ \frac{2\sin(\theta)\cos(\phi)-2\sin(\theta)}{2\cos(\theta)\cos(\phi)-2\cos\theta}$ $=\ \frac{2\sin\theta(\cos\phi-1)}{2\cos\theta(\cos\phi-1)}$ $=\ \frac{\sin\theta}{\cos\theta}=\tan\theta$ $=\ \text{RHS}$ $\therefore\ \frac{\sin(\theta+\phi)-2\sin\theta+\sin(\theta-\phi)}{\cos(\theta+\phi)-2\cos\theta+\cos(\theta-\phi)}=\tan\theta$ Hence proved.
View full question & answer→Question 264 Marks
$\text{If}\ \text{x}\cos\theta=\text{y}\cos\Big(\theta+\frac{2\pi}{3}\Big)=\text{z}\cos\Big(\theta+\frac{4\pi}{3}\Big),$ prove that $\text{xy}+\text{yz}+\text{zx}=0.$
AnswerGiven $\text{x}\cos\theta=\text{y}\cos\Big(\theta+\frac{2\pi}{3}\Big)=\text{z}\cos\Big(\theta+\frac{4\pi}{3}\Big)=\text{k}(\text{say})$ $\text{x}=\frac{\text{k}}{\cos\theta}$ $\text{y}=\frac{\text{k}}{\cos\Big(\theta+\frac{2\pi}{3}\Big)}$ $\text{z}=\frac{\text{k}}{\cos\Big(\theta+\frac{4\pi}{3}\Big)}$ $\text{xy}+\text{yz}+\text{zx}=\text{k}^2\Bigg[\frac{1}{\cos\theta\cos\big(\theta+\frac{2\pi}{3}\big)}+\frac{1}{\cos\big(\theta+\frac{2\pi}{3}\big)\cos\big(\theta+\frac{4\pi}{3}\big)}+\frac{1}{\cos\big(\theta+\frac{4\pi}{3}\big)\cos\theta}\Bigg]$ $=\ \text{k}^2\Bigg[\frac{\cos\big(\theta+\frac{4\pi}{3}\big)+\cos\theta+\cos\big(\theta+\frac{2\pi}{3}\big)}{\cos\theta\cos\big(\theta+\frac{2\pi}{3}\big)\cos\big(\theta+\frac{4\pi}{3}\big)}\Bigg]$ $=\ \text{k}^2\Bigg[\frac{\cos\theta\cos\frac{4\pi}{3}-\sin\theta\sin\frac{4\pi}{3}+\cos\theta+\cos\theta\cos\frac{2\pi}{3}-\sin\theta\sin\frac{2\pi}{3}}{\cos\theta\cos\big(\theta+\frac{2\pi}{3}\big)\cos\big(\theta+\frac{4\pi}{3}\big)}\Bigg]$ $=\ \text{k}^2\Bigg[\frac{\cos\theta\big(\frac{-1}{2}\big)-\sin\theta\big(\frac{-\sqrt3}{2}\big)+\cos\theta+\cos\theta\big(\frac{-1}{2}\big)-\sin\theta\big(\frac{\sqrt3}{2}\big)}{\cos\theta\cos\big(\theta+\frac{2\pi}{3}\big)\cos\big(\theta+\frac{4\pi}{3}\big)}\Bigg]$ $=\ \text{k}^2\Bigg[\frac{-\cos\theta+\sin\theta\big(\frac{\sqrt3}{2}\big)+\cos\theta+-\sin\theta\big(\frac{\sqrt3}{2}\big)}{\cos\theta\cos\big(\theta+\frac{2\pi}{3}\big)\cos\big(\theta+\frac{4\pi}{3}\big)}\Bigg]$ $=\ 0$ Hence proved.
View full question & answer→Question 274 Marks
Prove that: $\cos10^\circ\cos30^\circ\cos50^\circ\cos70^\circ=\frac{3}{16}$
Answer$\cos10^\circ\cos30^\circ\cos50^\circ\cos70^\circ=\frac{3}{16}$ $\text{LHS}=\cos10^\circ\cos30^\circ\cos50^\circ\cos70^\circ$ $=\ \cos30^\circ\cos10^\circ\cos50^\circ\cos70^\circ$ $=\ \frac{\sqrt3}{2}(\cos10^\circ\cos50^\circ\cos70^\circ)$ $=\ \frac{\sqrt3}{2}(\cos10^\circ\cos50^\circ)\cos70^\circ$ $=\ \frac{\sqrt3}{2}(2\cos10^\circ\cos50^\circ)\cos70^\circ$$[\text{Multiplying and dividing by 2}]$ Also, $\Rightarrow\ 2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})\ \dots(\text{i})$ $=\ \frac{\sqrt3}{4}\cos70^\circ(\cos(50^\circ+10^\circ)+\cos(10^\circ-50^\circ))$ $=\ \frac{\sqrt3}{4}\cos70^\circ(\cos60^\circ+(-40^\circ))$ Now, $\cos(-\theta)=\cos\theta$ $=\ \frac{\sqrt3}{4}\cos70^\circ\Big(\frac{1}{2}+\cos40^\circ\Big)\Big[\because\cos60^\circ=\frac{1}{2}\Big]$ $=\ \frac{\sqrt3}{8}\cos70^\circ+\frac{\sqrt3}{4}\cos70^\circ\cos40^\circ$ $=\ \frac{\sqrt3}{8}\cos70^\circ+\frac{\sqrt3}{8}(2\cos70^\circ\cos40^\circ)$ $=\ \frac{\sqrt3}{8}[\cos70^\circ+\cos(70^\circ+40^\circ)+\cos(70^\circ-40^\circ)][\text{from(i)}]$ $=\ \frac{\sqrt3}{8}[\cos70^\circ+\cos110^\circ+\cos30^\circ]$ $=\ \frac{\sqrt3}{8}\Big[\cos70^\circ+\cos(180^\circ-70^\circ)+\frac{\sqrt3}{2}\Big]$ $=\ \frac{\sqrt3}{8}\Big[\cos70^\circ-\cos70^\circ+\frac{\sqrt3}{2}\Big][\because\cos(180^\circ-\theta)=-\cos\theta]$ $=\ \frac{\sqrt3}{8}\times\frac{\sqrt3}{2}=\frac{3}{16}$ $=\ \text{RHS}$
View full question & answer→Question 284 Marks
Prove that: $\tan20^\circ\tan30^\circ\tan40^\circ\tan80^\circ=1$
Answer$\text{LHS}=\tan20^\circ\tan30^\circ\tan40^\circ\tan80^\circ$ $\frac{1}{\sqrt3}(\tan20^\circ\tan40^\circ\tan80^\circ)$$\Big[\because\ \tan30^\circ=\frac{1}{\sqrt3}\Big]$ $=\ \frac{(\sin20^\circ\sin40^\circ\sin80^\circ)}{(\cos20^\circ\cos40^\circ\cos80^\circ)\sqrt3}$ $=\ \frac{(2\sin20^\circ\sin40^\circ)\sin80^\circ}{\sqrt3(2\cos20^\circ\cos40^\circ)\cos80^\circ}$ Applying $\Rightarrow\ 2\sin\text{A}\sin\text{B}=\cos(\text{A}-\text{B})-\cos(\text{A+B})$ $2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})$ $=\ \frac{(\cos(40^\circ-20^\circ)-\cos(20^\circ+40^\circ))\sin80^\circ}{\cos(20^\circ+40^\circ)+\cos(40^\circ-20^\circ)\cos80^\circ\sqrt3}$ $=\ \frac{(\cos20^\circ-\cos60^\circ)\sin80^\circ}{\sqrt3(\cos60^\circ+\cos20^\circ)\cos80^\circ}$ $=\ \frac{\Big(\cos20^\circ-\frac{1}{2}\Big)\sin80^\circ}{\sqrt3\Big(\frac{1}{2}+\cos20^\circ\Big)\cos80^\circ}$ $=\ \frac{2\sin20^\circ\sin80^\circ-\sin80^\circ}{\sqrt3(\cos80^\circ+2\cos20^\circ\cos80^\circ)}$ Now, $\Rightarrow\ 2\sin\text{A}\cos\text{B}=\sin(\text{A+B})+\sin(\text{A}-\text{B})$ $=\ \frac{\sin(80^\circ+20^\circ)+\sin(80^\circ-20^\circ)-\sin80^\circ}{\sqrt3(\cos80^\circ+\cos(20^\circ+80^\circ)+\cos(80^\circ-20^\circ))}$ $=\ \frac{\sin100^\circ+\sin60^\circ-\sin80^\circ}{\sqrt3(\cos80^\circ+\cos100^\circ+\cos60^\circ)}$ $=\ \frac{\sin100^\circ+\sin60^\circ-\sin(80^\circ-100^\circ)}{\sqrt3(\cos80^\circ+\cos(1800^\circ-80^\circ)+\sin60^\circ)}$ $=\ \frac{\sin100^\circ+\frac{\sqrt3}{2}-\sin100^\circ}{\sqrt3(\cos80^\circ-\cos80^\circ+\cos60^\circ)}$ $=\ \frac{\frac{\sqrt3}{2}}{\sqrt3\big(\frac{1}{2}\big)}=1=\ \text{RHS}$
View full question & answer→Question 294 Marks
Prove that: $\cos\text{x}\cos\frac{\text{x}}{2}-\cos3\text{x}\cos\frac{9\text{x}}{2}=\sin7\text{x}\sin8\text{x}.$
AnswerConsider the left hand side of the given expression: $\text{LHS}=\cos\text{x}\cos\frac{\text{x}}{2}-\cos{3\text{x}}\cos\frac{9\text{x}}{2}$ We know that $2\cos\text{A}\cos\text{B}=\cos\text{A+B}+\cos(\text{A}-\text{B})$ Thus, $\text{LHS}=\frac{1}{2}\Big[\cos\Big(\text{x}+\frac{\text{x}}{2}\Big)+\cos\Big(\text{x}-\frac{\text{x}}{2}\Big)-\frac{1}{2}\Big[\cos\Big(3\text{x}+\frac{9\text{x}}{2}\Big)+\cos\Big(3\text{x}-\frac{9\text{x}}{2}\Big)\Big]$ $=\ \frac{1}{2}\Big[\cos\Big(\frac{3\text{x}}{2}\Big)+\cos\Big(\frac{\text{x}}{2}\Big)\Big]-\frac{1}{2}\Big[\cos\Big(\frac{15\text{x}}{2}\Big)+\cos\Big(-\frac{3\text{x}}{2}\Big)\Big]$ $=\ \frac{1}{2}\Big[\cos\Big(\frac{3\text{x}}{2}\Big)-\cos\Big(\frac{\text{x}}{2}\Big)\Big]-\frac{1}{2}\Big[\cos\Big(\frac{15\text{x}}{2}\Big)+\cos\Big(-\frac{3\text{x}}{2}\Big)\Big]$$[\because\ \cos(-\theta)=\cos\theta]$ $=\ \frac{1}{2}\Big[\cos\Big(\frac{3\text{x}}{2}\Big)+\cos\Big(\frac{\text{x}}{2}\Big)-\cos\Big(\frac{15\text{x}}{2}\Big)-\cos\Big(\frac{3\text{x}}{2}\Big)\Big]$ $=\ \frac{1}{2}\Big[\cos\Big(\frac{\text{x}}{2}\Big)-\cos\Big(\frac{15\text{x}}{2}\Big)\Big]$ Also we know that, $\cos\text{D}-\cos\text{C}=2\sin\frac{\text{C+D}}{2}\sin\frac{\text{C}-\text{D}}{2}$ Therefore, $\text{LHS}=\frac{1}{2}\times2\sin\frac{\frac{15\text{x}}{2}+\frac{\text{x}}{2}}{2}\sin\frac{\frac{15\text{x}}{2}-\frac{\text{x}}{2}}{2}$ $=\ \sin\frac{\frac{16\text{x}}{2}}{2}\sin\frac{\frac{14\text{x}}{2}}{2}$ $=\ \sin\frac{8\text{x}}{2}\sin\frac{7\text{x}}{2}$ $=\ \text{RHS}$ Note: Question given in the book is incorrect. RHS should be equal to $\sin\frac{8\text{x}}{2}\sin\frac{7\text{x}}{2}.$
View full question & answer→Question 304 Marks
$\text{If}\ \text{m}\sin\theta=\text{n}\sin(\theta+2\alpha),$ prove that $\tan(\theta+\alpha)\cot\alpha=\frac{\text{m+n}}{\text{m}-\text{n}}.$
AnswerGiven that $\text{m}\sin\theta=\text{n}\sin(\theta+2\alpha),$ We need to prove that $\tan(\theta+\alpha)=\frac{\text{m+n}}{\text{m}-\text{n}}\tan\alpha$ $\text{m}\sin\theta=\text{n}\sin(\theta+2\alpha)$ $\Rightarrow\ \frac{\sin(\theta+2\alpha)}{\sin\theta}=\frac{\text{m}}{\text{n}}$ Using Componendo - Dividendo, we have, $\Rightarrow\ \frac{\sin(\theta+2\alpha)+\sin\theta}{\sin(\theta+2\theta)-\sin\theta}=\frac{\text{m+n}}{\text{m}-\text{n}}...(1)$ We know that, $\sin\text{C}+\sin\text{D}=2\sin\frac{\text{C+D}}{2}\cos\frac{\text{C}-\text{D}}{2}$ and $\sin\text{C}-\sin\text{D}=2\cos\frac{\text{C+D}}{2}\sin\frac{\text{C}-\text{D}}{2}$ Applying the above formula in equation (1), we have, $\frac{2\sin\frac{\theta+2\theta+\theta}{2}\cos\frac{\theta+2\theta-\theta}{2}}{2\cos\frac{\theta+2\theta+\theta}{2}\sin\frac{\theta+2\theta-\theta}{2}}=\frac{\text{m+n}}{\text{m}-\text{n}}$ $\Rightarrow\ \frac{2\sin(\theta+\alpha)\cos\alpha}{2\cos(\theta+\alpha)\sin\alpha}=\frac{\text{m+n}}{\text{m}-\text{n}}$ $\Rightarrow\ \frac{\tan(\theta+\alpha)}{\tan\alpha}=\frac{\text{m+n}}{\text{m}-\text{n}}$ $\Rightarrow\ \tan(\theta+\alpha)=\frac{\text{m+n}}{\text{m}-\text{n}}\times\tan\alpha$ Hence proved.
View full question & answer→Question 314 Marks
Show that:$\sin\text{A}\sin(\text{B}-\text{C})+\sin\text{B}\sin(\text{C}-\text{A})+\sin\text{C}\sin(\text{A}-\text{B})=0$
AnswerWe have, $\text{LHS}=\ \sin\text{A}\sin(\text{B}-\text{C})+\sin\text{B}\sin(\text{C}-\text{A})+\sin\text{C}\sin(\text{A}-\text{B})$ $=\ \frac{1}{2}[2\sin\text{A}\sin(\text{B}-\text{C})+2\sin\text{B}\sin(\text{C}-\text{A})+2\sin\text{C}\sin(\text{A}-\text{B})]$ $=\ \frac{1}{2}\big[\cos(\text{A}-\text{B+C})-\cos(\text{A+B}-\text{C})+\cos(\text{B}-\text{C+A})\\\ \ \ \ \ \ \ \ -\cos(\text{B+C}-\text{A})+\cos(\text{C}-\text{A+B})-\cos(\text{C+A}-\text{B})\big]$ $=\ \frac{1}{2}\big[\cos(\text{A}-\text{B+C})-\cos(\text{A}-\text{B+C})-\cos(\text{A+B}-\text{C})\\\ \ \ \ \ \ \ \ +\cos(\text{A+B}-\text{C})-\cos(\text{B+C}-\text{A})+\cos(\text{B+C}-\text{A})\big]$ $=\ \frac{1}{2}\times0$ $=\ 0$ $=\ \text{RHS}$ $\therefore\ \sin\text{A}\sin(\text{B}-\text{C})+\sin\text{B}\sin(\text{C}-\text{A})+\sin\text{C}\sin(\text{A}-\text{B})$ $=0$ Hence proved.
View full question & answer→Question 324 Marks
Prove that: $\frac{\sin5\text{A}-\sin7\text{A}+\sin8\text{A}-\sin4\text{A}}{\cos4\text{A}+\cos7\text{A}-\cos5\text{A}-\cos8\text{A}}=\cot6\text{A}$
AnswerWe have, $\text{LHS}=\frac{\sin5\text{A}-\sin7\text{A}+\sin8\text{A}-\sin4\text{A}}{\cos4\text{A}+\cos7\text{A}-\cos5\text{A}-\cos8\text{A}}$ $=\ \frac{-(\sin7\text{A}-\sin5\text{A})+(\sin8\text{A}-\sin4\text{A})}{-(\cos7\text{A}-\cos5\text{A})-(\cos8\text{A}-\cos4\text{A})}$ $=\ \frac{-\Big[2\sin\Big(\frac{7\text{A}-5\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}+5\text{A}}{2}\Big)\Big]+\Big[2\sin\Big(\frac{8\text{A}-4\text{A}}{2}\Big)\cos\Big(\frac{8\text{A}+4\text{A}}{2}\Big)\Big]}{-2\sin\Big(\frac{7\text{A}+5\text{A}}{2}\Big)\sin\Big(\frac{7\text{A}-5\text{A}}{2}\Big)-\Big[-2\sin\Big(\frac{8\text{A}+4\text{A}}{2}\Big)\sin\Big(\frac{8\text{A}-4\text{A}}{2}\Big)\Big]}$ $=\ \frac{-2\sin\text{A}\cos6\text{A}+2\sin2\text{A}\cos6\text{A}}{-2\sin6\text{A}\sin\text{A}+2\sin6\text{A}\sin2\text{A}}$ $=\ \frac{2\cos6\text{A}[-\sin\text{A}+\sin2\text{A}]}{-2\sin6\text{A}[-\sin\text{A}+\sin2\text{A}]}$ $=\ \frac{\cos6\text{A}}{\sin6\text{A}}$ $=\ \cot6\text{A}$ $=\ \text{RHS}$ $\frac{\sin5\text{A}-\sin7\text{A}+\sin8\text{A}-\sin4\text{A}}{\cos4\text{A}+\cos7\text{A}-\cos5\text{A}-\cos8\text{A}}=\cot6\text{A}$ Hence proved.
View full question & answer→Question 334 Marks
Prove that: $\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ=\frac{3}{16}$
Answer$\text{LHS}=\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ$ $\sin20^\circ\sin40^\circ\sin80^\circ\times\frac{\sqrt3}{2}$$\Big[\because\ \sin60^\circ=\frac{\sqrt3}{2}\Big]$ $=\ \frac{\sqrt3}{2}\times\frac{1}{2}(2\sin20^\circ\sin40^\circ)\sin80^\circ$ $=\ \frac{\sqrt3}{4}[\cos(40^\circ-20^\circ)-\cos(40^\circ+20^\circ)]\sin80^\circ$ $=\ \frac{\sqrt3}{4}[\cos20^\circ-\cos60^\circ]\sin80^\circ$ $=\ \frac{\sqrt3}{4}\Big[\cos20^\circ\sin80^\circ-\frac{1}{2}\sin80^\circ\Big]$ $=\ \frac{\sqrt3}{8}[2\cos20^\circ\sin80^\circ-\sin80^\circ]$ $=\ \frac{\sqrt3}{8}[\sin(80^\circ+20^\circ)+\sin(80^\circ-20^\circ)-\sin80^\circ]$ $=\ \frac{\sqrt3}{8}[\sin100^\circ+\sin60^\circ-\sin80^\circ]$ $=\ \frac{\sqrt3}{8}[\sin80^\circ+\sin60^\circ-\sin80^\circ]$ $=\ \frac{\sqrt3}{8}\times\sin60^\circ=\frac{\sqrt3}{8}\times\frac{\sqrt3}{2}$ $=\ \frac{3}{16}=\text{RHS}$
View full question & answer→Question 344 Marks
Prove that: $\cos40^\circ\cos80^\circ\cos160=-\frac{1}{8}$
Answer$\cos40^\circ\cos80^\circ\cos160^\circ=-\frac{1}{8}$ $\text{LHS}=\cos40^\circ\cos80^\circ\cos160^\circ$ $=\ \cos80^\circ\cos40^\circ\cos160^\circ$ Multiplying and dividing by 2 $=\ \frac{1}{2}(\cos80^\circ\times(2\cos40^\circ\cos160^\circ))$ $2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})$ $=\ \frac{1}{2}(\cos80^\circ(\cos(40^\circ+160^\circ)+\cos(40^\circ-160^\circ)))$ $=\ \frac{1}{2}(\cos80^\circ(\cos200+\cos(-120)))$ $=\ \frac{1}{2}(\cos80^\circ(\cos180^\circ+20^\circ)+\cos(180^\circ-60^\circ))$ $=\ \frac{1}{2}\cos80^\circ(\cos20^\circ+\cos60^\circ)$ $=\ \frac{1}{2}\cos80^\circ\cos20^\circ+\frac{1}{2}\cos80^\circ+60^\circ$ $=\ -\frac{1}{2}(2\cos80^\circ\cos20^\circ)+\frac{1}{2}\cos80^\circ\cos60^\circ$ $=\ -\frac{1}{4}[2\cos80^\circ\cos20^\circ+\cos80^\circ]$ $=\ -\frac{1}{4}[\cos(80^\circ+20^\circ)+\cos(80^\circ-20^\circ)+\cos80^\circ]$ $=\ -\frac{1}{4}[\cos100^\circ+\cos60^\circ+\cos80^\circ]$ $=\ -\frac{1}{4}[\cos(180^\circ-80^\circ)+\cos60^\circ+\cos80^\circ]$ $=\ -\frac{1}{4}[-\cos80^\circ+\cos60^\circ+\cos80^\circ]$ $=\ -\frac{1}{4}\cos60^\circ$ $=\ -\frac{1}{4}\times\frac{1}{2}$ $=\ -\frac{1}{8}\ \text{RHS}$
View full question & answer→Question 354 Marks
Prove that: $\sin3\text{A}+\sin2\text{A}-\sin\text{A}=4\sin\text{A}\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}$
AnswerWe have, $\text{LHS}=\sin3\text{A}+\sin2\text{A}-\sin\text{A}$ $=\ \sin3\text{A}-\sin\text{A}+\sin2\text{A}$ $=\ 2\sin\Big(\frac{3\text{A}-\text{A}}{2}\Big)\cos\Big(\frac{3\text{A}+\text{A}}{2}\Big)+\sin2\text{A}$ $=\ 2\sin\text{A}\cos2\text{A}+\sin2\text{A}$ $=\ 2\sin\text{A}\cos2\text{A}+2\sin\text{A}\cos\text{A}$ $=\ 2\sin\text{A}[\cos2\text{A}+\cos\text{A}]$ $=\ 2\sin\text{A}\Big[2\cos\Big(\frac{2\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{2\text{A}-\text{A}}{2}\Big)\Big]$ $=\ 4\sin\text{A}\cos\frac{3\text{A}}{2}\cos\frac{\text{A}}{2}$ $=\ 4\sin\text{A}\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}$ $=\ \text{RHS}$ $\therefore\ \sin3\text{A}+\sin2\text{A}-\sin\text{A}=4\sin\text{A}\cos\frac{\text{A}}{2}\cos\frac{3\text{A}}{2}.$ Hence proved.
View full question & answer→Question 364 Marks
Prove that: $\frac{\sin\text{A}+2\sin3\text{A}+\sin5\text{A}}{\sin3\text{A}+2\sin5\text{A}+\sin7\text{A}}=\frac{\sin3\text{A}}{\sin5\text{A}}$
AnswerWe have,$\text{LHS}=\frac{\sin\text{A}+2\sin3\text{A}+\sin5\text{A}}{\sin3\text{A}+2\sin5\text{A}+\sin7\text{A}}$
$=\ \frac{\sin5\text{A}+\sin\text{A}+2\sin3\text{A}}{\sin7\text{A}+\sin3\text{A}+2\sin5\text{A}}$
$=\ \frac{2\sin\Big(\frac{5\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}-\text{A}}{2}\Big)+2\sin3\text{A}}{2\sin\Big(\frac{7\text{A}+3\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}-3\text{A}}{2}\Big)+2\sin5\text{A}}$
$=\ \frac{2\sin3\text{A}\cos2\text{A}+2\sin3\text{A}}{2\sin5\text{A}\cos2\text{A}+2\sin5\text{A}}$
$=\ \frac{2\sin3\text{A}(\cos2\text{A}+1)}{2\sin5\text{A}(\cos2\text{A}+1)}$
$=\ \frac{\sin3\text{A}}{\sin5\text{A}}$
$=\ \text{RHS}$
$\frac{\sin\text{A}+2\sin3\text{A}+\sin5\text{A}}{\sin3\text{A}+2\sin5\text{A}+\sin7\text{A}}=\frac{\sin3\text{A}}{\sin5\text{A}}$ Hence proved.
View full question & answer→Question 374 Marks
$\text{If}\ \text{y}\sin\phi=\text{x}\sin(2\theta+\phi),$ prove that $(\text{x+y})\cot(\theta+\phi)=(\text{y}-\text{x})\cot\theta$
AnswerWe have, $\text{y}\sin\phi=\text{x}\sin(2\theta+\phi)$ $\Rightarrow\ \frac{\sin\phi}{\sin(2\theta+\phi)}=\frac{\text{x}}{\text{y}}...(\text{i})$ Now, $\frac{\sin\phi}{\sin(2\theta+\phi)}=\frac{\text{x}}{\text{y}}$ $\Rightarrow\ \frac{\sin\phi}{\sin(2\theta+\phi)}+1=\frac{\text{x}}{\text{y}}+1$ $\Rightarrow\ \frac{\sin\phi+\sin(2\theta+\phi)}{\sin(2\theta+\phi)}=\frac{\text{x+y}}{\text{y}}...(\text{ii})$ Again, $\frac{\sin\phi}{\sin(2\theta+\phi)}=\frac{\text{x}}{\text{y}}$ [By equation (i)] $\Rightarrow\ \frac{\sin\phi}{\sin(2\theta+\phi)}-1=\frac{\text{x}}{\text{y}}-1$ $\Rightarrow\ \frac{\sin\phi-\sin(2\theta+\phi)}{\sin(2\theta+\phi)}=\frac{\text{x}-\text{y}}{\text{y}}...(\text{iii})$ Dividing equation (ii) by equation (iii), we get $\frac{\sin\phi+\sin(2\theta+\phi)}{\sin\phi-\sin(2\theta+\phi)}=\frac{\text{x+y}}{\text{x}-\text{y}}$ $\Rightarrow\ \frac{2\sin\big(\frac{\phi+2\theta+\phi}{2}\big)\cos\big(\frac{\phi-2\theta-\phi}{2}\big)}{2\sin\big(\frac{\phi-2\theta-\phi}{2}\big)\cos\big(\frac{\phi+2\theta+\phi}{2}\big)}=\frac{\text{x+y}}{\text{x}-\text{y}}$ $\Rightarrow\ \frac{\sin(\theta+\phi)\cos(\theta-\phi)}{\sin(-\theta)\cos(\theta+\phi)}=\frac{\text{x+y}}{\text{x}-\text{y}}$ $\Rightarrow\ \frac{\sin(\theta+\phi)\cos(\theta)}{\cos(\theta+\phi)[-\sin(\theta)]}=\frac{\text{x+y}}{\text{x}-\text{y}}$ $\Rightarrow\ \frac{-\cot(\theta)}{\cot(\theta+\phi)}=\frac{\text{x+y}}{\text{x}-\text{y}}$ $\Rightarrow\ -(\text{x}-\text{y})\cot\theta=(\text{x+y})\cot(\theta+\phi)$ $\Rightarrow\ (\text{y}-\text{x})\cot\theta=(\text{x+y})\cot(\theta+\phi)$ $\Rightarrow\ (\text{x}+\text{y})\cot(\theta+\phi)=(\text{y}-\text{x})\cot\theta$ Hence proved.
View full question & answer→Question 384 Marks
Prove that: $\frac{\sin\text{A}+\sin3\text{A}+\sin5\text{A}}{\cos\text{A}+\cos3\text{A}+\cos5\text{A}}=\tan3\text{A}$
AnswerWe have, $\text{LHS}=\frac{\sin\text{A}+\sin3\text{A}+\sin5\text{A}}{\cos\text{A}+\cos3\text{A}+\cos5\text{A}}$ $=\ \frac{(\sin5\text{A}+\sin\text{A})+\sin3\text{A}}{(\cos5\text{A}+\cos\text{A})+\cos3\text{A}}$ $=\ \frac{2\sin\Big(\frac{5\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}-\text{A}}{2}\Big)+\sin3\text{A}}{2\cos\Big(\frac{5\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}-\text{A}}{2}\Big)+\cos3\text{A}}$ $=\ \frac{2\sin3\text{A}\cos2\text{A}+\sin3\text{A}}{2\cos3\text{A}+\cos2\text{A}+\cos3\text{A}}$ $=\ \frac{\sin3\text{A}(2\cos2\text{A}+1)}{\cos3\text{A}(2\cos\text{A}+1)}$ $=\ \frac{\sin3\text{A}}{\cos3\text{A}}$ $=\ \tan3\text{A}$ $=\ \text{RHS}$ $\therefore\ \frac{\sin\text{A}+\sin3\text{A}+\sin5\text{A}}{\cos\text{A}+\cos3\text{A}+\cos5\text{A}}=\tan3\text{A}$ Hence proved.
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$\text{If}\ \sin2\text{A}=\lambda\sin2\text{B},$ prove that: $\frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}=\frac{\lambda+1}{\lambda-1}$
AnswerWe have, $\sin2\text{A}=\lambda\sin2\text{B}$ $\Rightarrow\ \lambda=\frac{\sin2\text{A}}{\sin2\text{B}}$ Now, $\frac{\lambda+1}{\lambda-1}=\frac{\frac{\sin2\text{A}}{\sin2\text{B}}+1}{\frac{\sin2\text{A}}{\sin2\text{B}}-1}$ $=\ \frac{\frac{\sin2\text{A}+\sin2\text{B}}{\sin2\text{B}}}{\frac{\sin2\text{A}-\sin2\text{B}}{\sin2\text{B}}}$ $=\ \frac{\sin2\text{A}+\sin2\text{B}}{\sin2\text{A}-\sin2\text{B}}$ $=\ \frac{2\sin\Big(\frac{2\text{A}+2\text{B}}{2}\Big)\cos\Big(\frac{2\text{A}-2\text{B}}{2}\Big)}{2\sin\Big(\frac{2\text{A}-2\text{B}}{2}\Big)\cos\Big(\frac{2\text{A}+2\text{B}}{2}\Big)}$ $=\ \frac{\sin(\text{A+B})\cos(\text{A}-\text{B})}{\sin(\text{A}-\text{B})\cos(\text{A+B})}$ $=\ \frac{\sin(\text{A+B})\cos(\text{A}-\text{B})}{\cos(\text{A}+\text{B})\sin(\text{A}-\text{B})}$ $=\ \frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}$ $\therefore\ \frac{\lambda+1}{\lambda-1}=\frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}$ $\Rightarrow\ \frac{\tan(\text{A+B})}{\tan(\text{A}-\text{B})}=\frac{\lambda+1}{\lambda-1}$ Hence proved.
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Prove that: $\frac{\cos3\text{A}+2\cos5\text{A}+\cos7\text{A}}{\cos\text{A}+2\cos3\text{A}+\cos5\text{A}}=\frac{\cos5\text{A}}{\cos3\text{A}}$
AnswerWe have, $\text{LHS}=\frac{\cos3\text{A}+2\cos5\text{A}+\cos7\text{A}}{2\cos\text{A}+2\cos3\text{A}+\cos5\text{A}}$ $=\ \frac{(\cos7\text{A}+\cos3\text{A})+2\cos5\text{A}}{(\cos5\text{A}+\cos\text{A})+2\cos3\text{A}}$ $=\ \frac{2\cos\Big(\frac{7\text{A}+3\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}-3\text{A}}{2}\Big)+2\cos5\text{A}}{2\cos\Big(\frac{5\text{A}+\text{A}}{2}\Big)\cos\Big(\frac{5\text{A}-\text{A}}{2}\Big)+\cos3\text{A}}$ $=\ \frac{2\cos5\text{A}\cos2\text{A}+2\cos5\text{A}}{2\cos3\text{A}\cos2\text{A}+2\cos3\text{A}}$ $=\ \frac{2\cos5\text{A}(\cos2\text{A}+1)}{2\cos3\text{A}(\cos2\text{A}+1)}$ $=\ \frac{\cos5\text{A}}{\cos3\text{A}}$ $=\ \tan3\text{A}$ $=\ \text{RHS}$ $\therefore\ \frac{\cos3\text{A}+2\sin5\text{A}+\cos7\text{A}}{\cos\text{A}+2\cos3\text{A}+\cos5\text{A}}=\frac{\cos5\text{A}}{\cos3\text{A}}$ Hence proved.
View full question & answer→Question 414 Marks
Prove that: $\cos3\text{A}+\cos5\text{A}+\cos7\text{A}+\cos15\text{A}=4\cos4\text{A}\cos5\text{A}\cos6\text{A}$
AnswerWe have, $\text{LHS}=\cos3\text{A}+\cos5\text{A}+\cos7\text{A}+\cos15\text{A}$ $=\ [\cos5\text{A}+\cos3\text{A}]+[\cos15\text{A}+\cos7\text{A}]$ $=\ \Big[2\cos\frac{5\text{A}+3\text{A}}{2}\cos\frac{5\text{A}-3\text{A}}{2}\Big]+\Big[2\cos\frac{15\text{A}+7\text{A}}{2}\cos\frac{15\text{A}-7\text{A}}{2}\Big]$ $=\ 2\cos4\text{A}\cos\text{A}+2\cos11\text{A}\cos4\text{A}$ $=\ 2\cos4\text{A}[\cos\text{A}+\cos11\text{A}]$ $=\ 2\cos4\text{A}[\cos11\text{A}+\cos\text{A}]$ $=\ 2\cos4\text{A}\Big[2\cos\frac{(11\text{A}+\text{A)}}{2}\cos\frac{(11\text{A}-\text{A})}{2}\Big]$ $=\ 4\cos\text{A}[\cos6\text{A}\cos5\text{A}]$ $=\ 4\cos4\text{A}\cos5\text{A}\cos6\text{A}$ $=\ \text{RHS}$ $\therefore\ \cos3\text{A}+\cos5\text{A}+\cos7\text{A}+\cos15\text{A}=4\cos4\text{A}\cos5\text{A}\cos6\text{A}$ Hence proved.
View full question & answer→Question 424 Marks
Prove that: $\frac{\sin\text{A}+\sin\text3{A}}{\cos\text{A}-\cos3\text{A}}=\cot\text{A}$
AnswerWe have, $\text{LHS}=\frac{\sin\text{A}+\sin3\text{A}}{\cos\text{A}-\cos3\text{A}}$ $=\ \frac{2\sin\Big(\frac{\text{A}+3\text{A}}{2}\Big)\cos\Big(\frac{\text{A}-3\text{A}}{2}}{-2\sin\Big(\frac{\text{A}+3\text{A}}{2}\Big)\sin\Big(\frac{\text{A}-3\text{A}}{2}\Big)}$ $=\ \frac{-\sin2\text{A}\times\cos(-\text{A})}{\sin2\text{A}\sin(-\text{A})}$ $=\ \frac{-\cos(-\text{A})}{\sin(-\text{A})}$ $=\ \frac{-\cos\text{A}}{-\sin\text{A}}$ $[\because\ \cos(-\theta)=\cos\theta\text{ and }\sin(-\theta)=-\sin\theta]$ $=\ \frac{\cos\text{A}}{\sin\text{A}}$ $=\ \cot\text{A}$ $=\ \text{RHS}$ $\therefore\ \frac{\sin\text{A}+\sin3\text{A}}{\cos\text{A}-\cos3\text{A}}=\cot\text{A}.$ Hence proved.
View full question & answer→Question 434 Marks
Prove that: $\sin\frac{\text{x}}{2}\sin\frac{7\text{x}}{2}+\sin\frac{3\text{x}}{2}\sin\frac{11\text{x}}{2}=\sin2\text{x}\sin5\text{x}.$
AnswerWe have, $\text{LHS}=\sin\frac{\text{x}}{2}\sin\frac{7\text{x}}{2}+\sin\frac{3\text{x}}{2}\sin\frac{11\text{x}}{2}$ $=\ \frac{1}{2}\Big[2\sin\frac{7\text{x}}{2}\sin\frac{\text{x}}{2}+2\sin\frac{110}{2}\sin\frac{3\text{x}}{2}\Big]$ $=\ \frac{1}{2}\Big[\cos\Big(\frac{7\text{x}}{2}-\frac{\text{x}}{2}\Big)-\cos\Big(\frac{7\text{x}}{2}+\frac{\text{x}}{2}\Big)+\cos\Big(\frac{110}{2}-\frac{3\text{x}}{2}\Big)-\cos\Big(\frac{110}{2}+\frac{3\text{x}}{2}\Big)\Big]$ $=\ \frac{1}{2}\Big[\cos\frac{60}{2}-\cos\frac{80}{2}+\cos\frac{80}{2}-\cos\frac{140}{2}\Big]$ $=\ \frac{1}{2}[\cos3\text{x}-\cos4\text{x}+\cos4\text{x}-\cos7\text{x}]$ $=\ \frac{1}{2}[\cos3\text{x}-\cos7\text{x}]$ $=\ \frac{-1}{2}[\cos7\text{x}-\cos3\text{x}]$ $=\ \frac{-1}{2}\Big[-2\sin\Big(\frac{7\text{x}+3\text{x}}{2}\Big)\sin\Big(\frac{7\text{x}-3\text{x}}{2}\Big)\Big]$ $=\ \sin\frac{10\text{x}}{2}\sin\frac{4\text{x}}{2}$ $=\ \sin5\text{x}\sin2\text{x}$ $=\ \sin2\text{x}\sin5\text{x}$ $=\ \text{RHS}$ $\therefore\ \sin\frac{\text{x}}{2}\sin\frac{7\text{x}}{2}+\sin\frac{3\text{x}}{2}\sin\frac{110}{2}=\sin2\text{x}\sin5\text{x}.$ Hence proved.
View full question & answer→Question 444 Marks
Prove that: $\frac{\sin3\text{A}\cos4\text{A}-\sin\text{A}\cos2\text{A}}{\sin4\text{A}\sin\text{A}+\cos6\text{A}\cos\text{A}}=\tan2\text{A}$
AnswerWe have, $\text{LHS}=\frac{\sin3\text{A}\cos4\text{A}-\sin\text{A}\cos2\text{A}}{\sin4\text{A}\sin\text{A}+\cos6\text{A}\cos\text{A}}$ $=\ \frac{2(\sin3\text{A}\cos4\text{A}-\sin\text{A}\cos2\text{A})}{2(\sin4\text{A}\sin\text{A}+\cos6\text{A}\cos\text{A})}$ $=\ \frac{2\sin3\text{A}\cos4\text{A}-2\sin\text{A}\cos2\text{A}}{2\sin4\text{A}\sin\text{A}+2\cos6\text{A}\cos\text{A}}$ $=\ \frac{\sin(4\text{A}+3\text{A})-\sin(4\text{A}-3\text{A})-[\sin(2\text{A}+\text{A})-\sin(2\text{A}-\text{A})]}{\cos(4\text{A}-\text{A})-\cos(4\text{A}+\text{A})+\cos(6\text{A}+\text{A})+\cos(6\text{A}-\text{A})}$ $=\ \frac{\sin(7\text{A})-\sin(\text{A})-\sin(3\text{A})+\sin(\text{A})}{\cos(3\text{A})-\cos(5\text{A})+\cos(7\text{A})+\cos(5\text{A})}$ $=\ \frac{\sin(7\text{A})-\sin(3\text{A})}{\cos(3\text{A})+\cos(7\text{A})}$ $=\ \frac{2\sin\Big(\frac{7\text{A}-3\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}+3\text{A}}{2}\Big)}{2\cos\Big(\frac{7\text{A}+3\text{A}}{2}\Big)\cos\Big(\frac{7\text{A}-3\text{A}}{2}\Big)}$ $=\ \frac{\sin2\text{A}}{\cos2\text{A}}$ $=\ \tan2\text{A}$ $=\ \text{RHS}$ $\therefore\ \frac{\sin3\text{A}\cos4\text{A}-\sin\text{A}\cos2\text{A}}{\sin4\text{A}\sin\text{A}+\cos6\text{A}\cos\text{A}}=\tan2\text{A}$ Hence proved.
View full question & answer→Question 454 Marks
Prove that: $\text{If}\cos\text{A}+\cos\text{B}=\frac{1}{2}\text{ and }\sin\text{A}+\sin\text{B}=\frac{1}{4},$ prove that $\tan\Big(\frac{\text{A+B}}{2}\Big)=\frac{1}{2}.$
AnswerWe have, $\text{LHS}=\cos\text{A}+\cos\text{B}=\frac{1}{2}$ $\text{and},\ \sin\text{A}+\sin\text{B}=\frac{1}{4}$ Now, $\frac{\sin\text{A}+\sin\text{B}}{\cos\text{A}+\cos\text{B}}=\frac{\frac{1}{4}}{\frac{1}{2}}$ [On dividing] $\Rightarrow\ \frac{2\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}{2\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)\cos\Big(\frac{\text{A}-\text{B}}{2}\Big)}=\frac{1}{2}$ $\Rightarrow\ \frac{\sin\Big(\frac{\text{A}+\text{B}}{2}\Big)}{\cos\Big(\frac{\text{A}+\text{B}}{2}\Big)}=\frac{1}{2}$ $\Rightarrow\ \tan\Big(\frac{\text{A}+\text{B}}{2}\Big)=\frac{1}{2}$ $\Rightarrow\ \text{RHS}$ Hence proved.
View full question & answer→Question 464 Marks
Prove that: $\frac{\cos4\text{A}+\cos3\text{A}+\cos2\text{A}}{\sin4\text{A}+\sin3\text{A}+\sin2\text{A}}=\cot3\text{A}$
AnswerWe have, $\text{LHS}=\frac{\cos4\text{A}+\cos3\text{A}+\cos2\text{A}}{\sin4\text{A}+\sin2\text{A}+\sin2\text{A}}$ $=\ \frac{(\cos4\text{A}+\cos2\text{A})+\cos3\text{A}}{(\sin4\text{A}+\sin2\text{A})+\sin3\text{A}}$ $=\ \frac{2\cos\Big(\frac{4\text{A}+2\text{A}}{2}\Big)\cos\Big(\frac{4\text{A}-2\text{A}}{2}\Big)+\cos3\text{A}}{2\sin\Big(\frac{4\text{A}+2\text{A}}{2}\Big)\cos\Big(\frac{4\text{A}-2\text{A}}{2}\Big)+\sin3\text{A}}$ $=\ \frac{2\cos3\text{A}\cos\text{A}+\cos3\text{A}}{2\sin3\text{A}\cos\text{A}+\sin3\text{A}}$ $=\ \frac{\cos3\text{A}(2\cos\text{A}+1)}{\sin3\text{A}(2\cos2\text{A}+1)}$ $=\ \frac{\cos3\text{A}}{\sin\text{A}}$ $=\ \cot3\text{A}$ $=\ \text{RHS}$ $\therefore\ \frac{\cos4\text{A}+\cos3\text{A}+\cos2\text{A}}{\sin4\text{A}+\sin3\text{A}+\sin2\text{A}}=\cot3\text{A}$ Hence proved.
View full question & answer→Question 474 Marks
Prove that: $\sin20^\circ\sin40^\circ\sin80^\circ=\frac{\sqrt3}{8}$
Answer$\text{LHS}=\sin20^\circ\sin40^\circ\sin80^\circ$ $=\ \frac{1}{2}(2\sin20^\circ\sin40^\circ)\sin80^\circ$ $=\ \frac{1}{2}[\cos(40^\circ-20^\circ)-\cos(40^\circ+20^\circ)]\sin80^\circ$$[\because2\sin\text{A}\sin\text{B}=\cos(\text{A}-\text{B})-\cos(\text{A+B})$ $=\ \frac{1}{2}[\cos20^\circ-\cos60^\circ]\sin80^\circ$ $=\ \frac{1}{2}\Big[\cos20^\circ-\frac{1}{2}\Big]\sin80^\circ$ $=\ \frac{1}{2}[\cos20^\circ\sin80^\circ]-\frac{1}{4}\sin80^\circ$ $=\ \frac{1}{4}[2\cos20^\circ\sin80^\circ-\sin80^\circ]$ $=\ \frac{1}{4}[\sin(80^\circ+20^\circ)+\sin(80^\circ-20^\circ)-\sin80^\circ]$$[\because2\sin\text{A}\cos\text{B}=\sin(\text{A+B})+\sin(\text{A}-\text{B})$ $=\ \frac{1}{4}[\sin100^\circ+\sin60^\circ-\sin80^\circ]$ $=\ \frac{1}{4}\Big[\sin(180^\circ-80^\circ)+\frac{\sqrt3}{2}-\sin80^\circ\Big]$ $=\ \frac{1}{4}\Big[\sin80^\circ+\frac{\sqrt3}{2}-\sin80^\circ\Big]$ $= \frac{\sqrt3}{8}=\ \text{RHS}$
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