Solve the Following Question.(5 Marks)

Question 15 Marks

Prove that:
$\cos20^\circ\cos100^\circ+\cos100^\circ\cos140^\circ-\cos140^\circ\cos200^\circ=-\frac{3}{4}$

Answer

We 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.

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Question 25 Marks

Prove that:
$\cos(\text{A+B+C})+\cos(\text{A}-\text{B+C})+\cos(\text{A+B}-\text{C})\\+\cos(-\text{A+B+C})=4\cos\text{A}\cos\text{B}\cos\text{C}$

Answer

We have,
$\text{LHS}=\cos(\text{A+B+C})+\cos(\text{A}-\text{B}+\text{C})\\ \ \ \ \ +\cos(\text{A+B}-\text{C})+\cos(-\text{A+B+C})$
$=\ [\cos(\text{A+B+C})+\cos(\text{A}-\text{B+C})]\\ \ \ \ \ +[\cos(\text{A+B}-\text{C})+\cos(-\text{A+B+C})$
$=\ 2\cos\Big\{\frac{\text{A+B+C+A}-\text{B}+\text{C}}{2}\Big\}\cos\Big\{\frac{\text{A+B+C}-\text{A+B}-\text{C}}{2}\Big\}\\ \ \ \ +2\begin{Bmatrix}\cos\Big\{\frac{\text{A+B}-\text{C}-\text{A+B+C}}{2}\Big\} \\\cos\Big\{\frac{\text{A+B}-\text{C}+\text{A}-\text{B}-\text{C}}{2}\Big\} \end{Bmatrix}$
$=\ 2\cos\Big(\frac{2\text{A}+2\text{C}}{2}\Big)\cos\Big(\frac{2\text{B}}{2}\Big)+2\cos\Big(\frac{2\text{B}}{2}\Big)\cos\Big\{\frac{2\text{A}-2\text{C}}{2}\Big\}$
$=\ 2\cos(\text{A+B})\cos(\text{B})+2\cos(\text{B})\cos(\text{A}-\text{C})$
$=\ 2\cos(\text{B})[\cos(\text{A+C})+\cos(\text{A}-\text{C})]$
$=\ 2\cos\text{B}\Big[2\cos\Big(\frac{\text{A+C+A}-\text{C}}{2}\Big)\cos\Big(\frac{\text{A+C}-\text{A+C}}{2}\Big)\Big]$
$=\ 2\cos(\text{B})[2\cos\text{A}\cos\text{C}]$
$=\ 4\cos\text{A}\cos\text{B}\cos{C}.$
$=\ \text{RHS}$

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Question 35 Marks

prove that:
$\sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+\sin(\text{C}-\text{A})\cos(\text{B}-\text{D})+\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})=0$

Answer

We 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.

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Question 45 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 $\tan\text{A}\tan\text{B}\tan\text{C}\tan\text{D}=-1$

Answer

We 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.

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Question 55 Marks

$\text{If}\ \cos(\alpha+\beta)\sin(\gamma+\delta)=\cos(\alpha-\beta)\sin(\gamma-\delta),$
prove that $\cot\alpha\cot\beta\cot\gamma=\cot\delta$

Answer

We 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.

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Question 65 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

$\frac{\pi}{3}=60^\circ$
$\text{LHS}=\tan\text{x}\tan(60^\circ-\text{x})\tan(60^\circ+\text{x})$
$=\ \frac{\sin\text{x}\sin(60^\circ-\text{x})\sin(60^\circ+\text{x})}{\cos\text{x}\cos(60^\circ-\text{x})\cos(60^\circ+\text{x})}$
$=\ \frac{\sin\text{x}(\sin^260^\circ-\sin^2\text{x})}{\cos\text{x}(\cos^260^\circ-\sin^2\text{x})}$
$=\ \frac{\sin\text{x}\Big(\frac{3}{4}-\sin^2\text{x}\Big)}{\cos\text{x}\Big(\frac{1}{4}-\sin^2\text{x}\Big)}$
$=\ \frac{\sin\text{x}(3-4\sin^2\text{x})}{\cos\text{x}(1-4\sin^2\text{x})}$
$=\ \frac{3\sin\text{x}-4\sin^3\text{x}}{4\cos^3\text{x}-3\cos\text{x}}$
$=\ \frac{\sin3\text{x}}{\cos3\text{x}}$
$=\ \tan3\text{x}=\text{RHS}$

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Question 75 Marks

prove that:
$\frac{\cos(\text{A+B+C})+\cos(-\text{A+B+C})+\cos(\text{A}-\text{B+C})+\cos(\text{A+B}-\text{C})}{\sin(\text{A+B+C})+\sin(-\text{A+B+C})+\sin(\text{A}-\text{B+C})-\sin(\text{A+B}-\text{C})}=\cot\text{C}$

Answer

We 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.

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Question 85 Marks

Prove that:
$\sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)\\=4\sin\Big(\frac{\alpha+\beta}{2}\Big)\sin\Big(\frac{\beta+\gamma}{2}\Big)\sin\Big(\frac{\gamma+\alpha}{2}\Big)$

Answer

We 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.

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Question 95 Marks

Prove that:
$\frac{\sin\text{A}\sin2\text{A}+\sin3\text{A}\sin6\text{A}}{\sin\text{A}\cos2\text{A}+\sin3\text{A}\cos6\text{A}}=\tan5\text{A}$

Answer

We 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.

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Question 105 Marks

Show that:$\sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+\sin(\text{C}-\text{A})\\\cos(\text{B}-\text{D})+\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})=0$

Answer

We 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.

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Question 115 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}. $

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Maths Solve the Following Question.(5 Marks)

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