Question 15 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 25 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})$
$=\ \frac{1}{4}[\sin100^\circ+\sin60^\circ-\sin80^\circ]$
$=\ \frac{1}{4}\Big[\sin(180^\circ-80^\circ)+\frac{\sqrt3}{2}-\sin80^\circ\Big]$
$=\ \frac{1}{4}\Big[\sin80^\circ+\frac{\sqrt3}{2}-\sin80^\circ\Big]$
$= \frac{\sqrt3}{8}=\ \text{RHS}$
View full question & answer→Question 35 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 45 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 55 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 65 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$$[\because\ \sin(-\theta)=-\sin\theta]$
$=\ 0$
$=\ \text{RHS}$
$\therefore\ \sin(\text{B}-\text{C})\cos(\text{A}-\text{D})+\sin(\text{C}-\text{A})\\\ \ \ \ \ \cos(\text{B}-\text{D})+\sin(\text{A}-\text{B})\cos(\text{C}-\text{D})=0$
Hence proved.
View full question & answer→Question 75 Marks
Prove that:
$\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ=\frac{3}{16}$
Answer$\text{LHS}=\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ$
$\sin20^\circ\sin40^\circ\sin80^\circ\times\frac{\sqrt3}{2}$$\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 85 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 95 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 105 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 115 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 125 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 135 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 145 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 155 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 165 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 175 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 185 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 195 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 205 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$$[\text{Multiplying and dividing by 2}]$
Also,
$\Rightarrow\ 2\cos\text{A}\cos\text{B}=\cos(\text{A+B})+\cos(\text{A}-\text{B})\ \dots(\text{i})$
$=\ \frac{\sqrt3}{4}\cos70^\circ(\cos(50^\circ+10^\circ)+\cos(10^\circ-50^\circ))$
$=\ \frac{\sqrt3}{4}\cos70^\circ(\cos60^\circ+(-40^\circ))$
Now,
$\cos(-\theta)=\cos\theta$
$=\ \frac{\sqrt3}{4}\cos70^\circ\Big(\frac{1}{2}+\cos40^\circ\Big)\Big[\because\cos60^\circ=\frac{1}{2}\Big]$
$=\ \frac{\sqrt3}{8}\cos70^\circ+\frac{\sqrt3}{4}\cos70^\circ\cos40^\circ$
$=\ \frac{\sqrt3}{8}\cos70^\circ+\frac{\sqrt3}{8}(2\cos70^\circ\cos40^\circ)$
$=\ \frac{\sqrt3}{8}[\cos70^\circ+\cos(70^\circ+40^\circ)+\cos(70^\circ-40^\circ)][\text{from(i)}]$
$=\ \frac{\sqrt3}{8}[\cos70^\circ+\cos110^\circ+\cos30^\circ]$
$=\ \frac{\sqrt3}{8}\Big[\cos70^\circ+\cos(180^\circ-70^\circ)+\frac{\sqrt3}{2}\Big]$
$=\ \frac{\sqrt3}{8}\Big[\cos70^\circ-\cos70^\circ+\frac{\sqrt3}{2}\Big][\because\cos(180^\circ-\theta)=-\cos\theta]$
$=\ \frac{\sqrt3}{8}\times\frac{\sqrt3}{2}=\frac{3}{16}$
$=\ \text{RHS}$
View full question & answer→Question 215 Marks
Prove that:
$\tan20^\circ\tan30^\circ\tan40^\circ\tan80^\circ=1$
Answer$\text{LHS}=\tan20^\circ\tan30^\circ\tan40^\circ\tan80^\circ$
$\frac{1}{\sqrt3}(\tan20^\circ\tan40^\circ\tan80^\circ)$$\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 225 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 235 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 245 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 255 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 265 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 275 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 285 Marks
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 295 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 305 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 315 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 325 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
$\frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{\cos(\alpha+\beta)-\cos(\alpha-\beta)}=\frac{\sin(\gamma-\delta)+\sin(\gamma+\delta)}{\sin(\gamma-\delta)-\sin(\gamma+\delta)}$
$\Rightarrow\ \frac{\cos(\alpha+\beta)+\cos(\alpha-\beta)}{\cos(\alpha+\beta)-\cos(\alpha-\beta)}=-\Big[\frac{\sin(\gamma+\delta)+\sin(\gamma-\delta)}{\sin(\gamma+\delta)-\sin(\gamma-\delta)}\Big]$
$\Rightarrow\ \frac{2\cos\big\{\frac{\alpha+\beta+\alpha-\beta}{2}\big\}\cos\big\{\frac{\alpha+\beta-\alpha+\beta}{2}\big\}}{-2\sin\big\{\frac{\alpha+\beta+\alpha-\beta}{2}\big\}\sin\big\{\frac{\alpha+\beta-\alpha+\beta}{2}\big\}}\\ \ \ \ \ =-\Bigg[\frac{2\sin\big\{\frac{\gamma+\delta+\gamma-\delta}{2}\big\}\cos\big\{\frac{\gamma+\delta-\gamma+\delta}{2}\big\}}{2\sin\big\{\frac{\gamma+\delta-\gamma+\delta}{2}\big\}\cos\big\{\frac{\gamma+\delta+\gamma-\delta}{2}\big\}}\Bigg]$
$$$\Rightarrow\ \frac{\cos\alpha\cos\beta}{\sin\alpha\sin\beta}=\frac{\sin\gamma\cos\delta}{\sin\delta\cos\gamma}$
$\Rightarrow\ \cot\alpha\cot\beta=\frac{\sin\gamma\cos\delta}{\cos\gamma\sin\delta}$
$\Rightarrow\ \cot\alpha\cot\beta=\frac{\cot\delta}{\cot\gamma}$
$\Rightarrow\ \cot\alpha\cot\beta\cot\gamma=\cot\delta$
$\therefore\ \cot\alpha\cot\beta\cot\gamma=\cot\delta$ Hence proved.
View full question & answer→Question 335 Marks
Prove that:
$\sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)\\=4\sin\Big(\frac{\alpha+\beta}{2}\Big)\sin\Big(\frac{\beta+\gamma}{2}\Big)\sin\Big(\frac{\gamma+\alpha}{2}\Big)$
AnswerWe have,
$\text{LHS}=\sin\alpha+\sin\beta+\sin\gamma-\sin(\alpha+\beta+\gamma)$
$=\ (\sin\alpha+\sin\beta)+(\sin\gamma-\sin(\alpha+\beta+\gamma))$
$=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)+2\sin\Big(\frac{\gamma-(\alpha+\beta+\gamma)}{2}\Big)\cos\Big(\frac{\gamma+\alpha+\beta+\gamma}{2}\Big)$
$=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)+2\sin\Big(\frac{-\alpha-\beta}{2}\Big)\cos\Big(\frac{\alpha+\beta+2\gamma}{2}\Big)$
$=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha-\beta}{2}\Big)-2\sin\Big(\frac{\alpha+\beta}{2}\Big)\cos\Big(\frac{\alpha+\beta+2\gamma}{2}\Big)$
$=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\Big[\cos\Big(\frac{\alpha-\beta}{2}\Big)-\cos\Big(\frac{\alpha+\beta+2\gamma}{2}\Big)\Big]$
$=\ 2\sin\Big(\frac{\alpha+\beta}{2}\Big)\Bigg[-2\sin\frac{\Big[\frac{\alpha-\beta}{2}+\frac{\alpha+\beta+2\gamma}{2}\Big]}{2}\sin\frac{\Big[\frac{\alpha-\beta}{2}-\frac{\alpha+\beta+2\gamma}{2}\Big]}{2}\Bigg]$
$$$=\ 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 345 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 355 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 365 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 375 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 385 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 395 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 405 Marks
$\text{If}\ \sin2\text{A}=\lambda\sin2\text{B},$ 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.
View full question & answer→Question 415 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 425 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 435 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.
View full question & answer→Question 445 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 455 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 465 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 475 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.
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