Question 13 Marks
$\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ}=1$
Answer\begin{aligned}
& \text { L.H.S. }=\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ} \\
& =\frac{\sin 6^{\circ}}{\cos 6^{\circ}} \cdot \frac{\sin 42^{\circ}}{\cos 42^{\circ}} \cdot \frac{\sin 66^{\circ}}{\cos 66^{\circ}} \cdot \frac{\sin 78^{\circ}}{\cos 78^{\circ}} \\
& =\frac{\left(2 \sin 66^{\circ} \sin 6^{\circ}\right)\left(2 \sin 78^{\circ} \sin 42^{\circ}\right)}{\left(2 \cos 66^{\circ} \cos 6^{\circ}\right)\left(2 \cos 78^{\circ} \cos 42^{\circ}\right)} \\
& =\frac{\cos \left(66^{\circ}-6^{\circ}\right)-\cos \left(66^{\circ}+6^{\circ}\right)}{\cos \left(66^{\circ}+6^{\circ}\right)+\cos \left(66^{\circ}-6^{\circ}\right)} \\
& . \frac{\cos \left(78^{\circ}-42^{\circ}\right)-\cos \left(78^{\circ}+42^{\circ}\right)}{\cos \left(78^{\circ}+42^{\circ}\right)+\cos \left(78^{\circ}-42^{\circ}\right)} \\
& =\frac{\left(\cos 60^{\circ}-\cos 72^{\circ}\right)\left(\cos 36^{\circ}-\cos 120^{\circ}\right)}{\left(\cos 60^{\circ}+\cos 72^{\circ}\right)\left(\cos 36^{\circ}+\cos 120^{\circ}\right)} \\
& =\frac{\left(\cos 60^{\circ}-\sin 18^{\circ}\right)\left(\cos 36^{\circ}+\sin 30^{\circ}\right)}{\left(\cos 60^{\circ}+\sin 18^{\circ}\right)\left(\cos 36^{\circ}-\sin 30^{\circ}\right)} \\
& \ldots\left[\because \cos \left(90^{\circ}+\theta\right)=-\sin \theta\right] \\
& =\frac{\left(\frac{1}{2}-\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}+\frac{1}{2}\right)}{\left(\frac{1}{2}+\frac{\sqrt{5}-1}{4}\right)\left(\frac{\sqrt{5}+1}{4}-\frac{1}{2}\right)} \\
& =\frac{9-5}{5-1}\\
& =1 \\
& =\text { R.H.S. } \\
\end{aligned}
View full question & answer→Question 23 Marks
$\sin \frac{\pi^c}{8}=\frac{1}{2} \sqrt{2-\sqrt{2}}$
AnswerWe know that $\cos \frac{\pi}{4}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}$
Also $\frac{\pi}{8}$ lies in the first quadrant, hence $\sin \frac{\pi}{8}$ is positive.
Now, $\cos 2 \theta=1-2 \sin ^2 \theta$
By putting $\theta=\frac{\pi}{8}$, we get,
$\begin{aligned}
& \cos \frac{\pi}{4}=1-2 \sin ^2 \frac{\pi}{8} \\
& \therefore 2 \sin ^2 \frac{\pi}{8}=1-\cos \frac{\pi}{4} \\
& =1-\frac{1}{\sqrt{2}} \\
& =\frac{\sqrt{2}-1}{\sqrt{2}} \\
& \therefore \sin ^2 \frac{\pi}{8}=\frac{\sqrt{2}-1}{2 \sqrt{2}} \\
& =\frac{\sqrt{2}(\sqrt{2}-1)}{4} \\
& \therefore \frac{2-\sqrt{2}}{4} \\
& \therefore \sin \frac{\pi}{8}=\frac{1}{\sqrt{2-\sqrt{2}}}
\end{aligned}$
View full question & answer→Question 33 Marks
$\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$
AnswerLet $\theta=18^{\circ}$
$\begin{aligned}
& \therefore 5 \theta=90^{\circ} \\
& \therefore 2 \theta+3 \theta=90^{\circ} \\
& \therefore 2 \theta=90^{\circ}-3 \theta \\
& \therefore \sin 2 \theta=\sin \left(90^{\circ}-3 \theta\right) \\
& \therefore \sin 2 \theta=\cos 3 \theta \\
& \therefore 2 \sin \theta \cos \theta=4 \cos ^3 \theta-3 \cos \theta \\
& \left.\therefore 2 \sin \theta=4 \cos ^2 \theta-3 \ldots . \because \cos \theta \neq 0\right] \\
& \therefore 2 \sin \theta=4\left(1-\sin ^2 \theta\right)-3 \\
& \therefore 2 \sin \theta=1-4 \sin ^2 \theta \\
& \therefore 4 \sin 2 \theta+2 \sin \theta-1=0 \\
& \therefore \sin \theta=\frac{-2 \pm \sqrt{4+16}}{8} \\
& =\frac{-2 \pm 2 \sqrt{5}}{8} \\
& =\frac{-1 \pm \sqrt{5}}{4}
\end{aligned}$
Since, $\sin 18^{\circ}>0$
$\therefore \sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$
View full question & answer→Question 43 Marks
$\sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}=\frac{\sqrt{3}}{8}$
Answer\begin{aligned}
& \text { L.H.S. }=\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 80^{\circ} \\
& =\sin 20^{\circ} \cdot \sin 40^{\circ} \cdot \sin 80^{\circ} \\
& =\frac{1}{2}\left(2 \cdot \sin 40^{\circ} \cdot \sin 20^{\circ}\right) \cdot \sin 80^{\circ} \\
& =\frac{1}{2}\left[\cos \left(40^{\circ}-20^{\circ}\right)-\cos \left(40^{\circ}+20^{\circ}\right)\right] \cdot \sin 80^{\circ} \\
& =\frac{1}{2}\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \sin 80^{\circ} \\
& =\frac{1}{2} \cdot \cos 20^{\circ} \cdot \sin 80^{\circ}-\frac{1}{2} \cdot \cos 60^{\circ} \cdot \sin 80^{\circ} \\
& =\frac{1}{2 \times 2}\left(2 \sin 80^{\circ} \cdot \cos 20^{\circ}\right)-\frac{1}{2 \times 2} \cdot \sin 80^{\circ} \\
& =\frac{1}{4}\left[\sin \left(80^{\circ}+20^{\circ}\right)+\sin \left(80^{\circ}-20^{\circ}\right)\right]-\frac{1}{2} \cdot \sin 80^{\circ} \\
& =\frac{1}{4} \cdot\left(\sin 100^{\circ}+\sin 60^{\circ}\right)-\frac{1}{4} \cdot \sin 80^{\circ} \\
& =\frac{1}{4} \sin 100^{\circ}+\frac{1}{4} \sin 60^{\circ}-\frac{1}{4} \sin 80^{\circ} \\
& =\frac{1}{4} \cdot \sin \left(180^{\circ}-80^{\circ}\right)+\frac{1}{4} \times \frac{\sqrt{3}}{2}-\frac{1}{4} \cdot \sin 80^{\circ} \\
& =\frac{1}{4} \sin 80^{\circ}+\frac{\sqrt{3}}{8}-\frac{1}{4} \sin 80^{\circ} \\
& \left.=\frac{\sqrt{3}}{8}=\text { R.H.S. } [\sin \left(180^{\circ}-\theta\right)=\sin \theta\right]
\end{aligned}
View full question & answer→Question 53 Marks
$3 \tan ^6 10^{\circ}-27 \tan ^4 10^{\circ}+33 \tan ^2 10^{\circ}=1$
AnswerSince, $\tan 30^{\circ}=\frac{1}{\sqrt{3}}$
$\begin{array}{ll}
\therefore \quad \tan 3\left(10^{\circ}\right)=\frac{1}{\sqrt{3}}\\
\therefore \quad & \frac{3 \tan 10^{\circ}-\tan ^3 10^{\circ}}{1-3 \tan ^2 10^{\circ}}=\frac{1}{\sqrt{3}}
\end{array}$
Squaring both sides, we get
$\begin{array}{ll}
& \frac{\left(3 \tan 10^{\circ}-\tan ^3 10^{\circ}\right)^2}{\left(1-3 \tan ^2 10^{\circ}\right)^2}=\frac{1}{3} \\
\therefore \quad & \frac{9 \tan ^2 10^{\circ}-6 \tan ^4 10^{\circ}+\tan ^6 10^{\circ}}{1-6 \tan ^2 10^{\circ}+9 \tan ^4 10^{\circ}}=\frac{1}{3} \\
\therefore \quad & 3\left(9 \tan ^2 10^{\circ}-6 \tan ^4 10^{\circ}+\tan ^6 10^{\circ}\right) \\
=1-6 \tan ^2 10^{\circ}+9 \tan ^4 10^{\circ} \\
\therefore \quad & 27 \tan ^2 10^{\circ}-18 \tan ^4 10^{\circ}+3 \tan ^6 10^{\circ} \\
=1-6 \tan ^2 10^{\circ}+9 \tan { }^4 10^{\circ} \\
\therefore \quad & 3 \tan ^6 10^{\circ}-27 \tan ^4 10^{\circ}+33 \tan ^2 10^{\circ}=1
\end{array}$
View full question & answer→Question 63 Marks
If $\sin 2 \mathrm{~A}=\lambda \sin 2 \mathrm{~B}$, then prove that $\frac{\tan (A+B)}{\tan (A-B)}=\frac{\lambda+1}{\lambda-1}$
Answer$\begin{aligned}
& \sin 2 A=\lambda \sin 2 \mathrm{~B} \\
& \therefore \quad \frac{\sin 2 A}{\sin 2 \mathrm{~B}}=\frac{\lambda}{1}
\end{aligned}$
By componendo-dividendo, we get
$\begin{array}{ll}
& \frac{\sin 2 \mathrm{~A}+\sin 2 \mathrm{~B}}{\sin 2 \mathrm{~A}-\sin 2 \mathrm{~B}}=\frac{\lambda+1}{\lambda-1} \\
\therefore \quad & \frac{2 \sin \left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right) \cos \left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)}{2 \cos \left(\frac{2 \mathrm{~A}+2 \mathrm{~B}}{2}\right) \sin \left(\frac{2 \mathrm{~A}-2 \mathrm{~B}}{2}\right)}=\frac{\lambda+1}{\lambda-1} \\
\therefore \quad & \frac{2 \sin (\mathrm{A}+\mathrm{B}) \cos (\mathrm{A}-\mathrm{B})}{2 \cos (\mathrm{A}+\mathrm{B}) \sin (\mathrm{A}-\mathrm{B})}=\frac{\lambda+1}{\lambda-1} \\
\therefore \quad & \tan (\mathrm{A}+\mathrm{B}) \cdot \cot (\mathrm{A}-\mathrm{B})=\frac{\lambda+1}{\lambda-1} \\
\therefore \quad & \frac{\tan (\mathrm{A}+\mathrm{B})}{\tan (\mathrm{A}-\mathrm{B})}=\frac{\lambda+1}{\lambda-1}
\end{array}$
View full question & answer→Question 73 Marks
$\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x}=\tan x$
Answer$\begin{aligned}
& \text { L.H.S. }=\frac{\sin 5 x-2 \sin 3 x+\sin x}{\cos 5 x-\cos x} \\
& =\frac{(\sin 5 x+\sin x)-2 \sin 3 x}{\cos 5 x-\cos x} \\
& =\frac{2 \sin \left(\frac{5 x+x}{2}\right) \cos \left(\frac{5 x-x}{2}\right)-2 \sin 3 x}{-2 \sin \left(\frac{5 x+x}{2}\right) \sin \left(\frac{5 x-x}{2}\right)} \\
& =\frac{2 \sin 3 x \cos 2 x-2 \sin 3 x}{-2 \sin 3 x \sin 2 x} \\
& =\frac{2 \sin 3 x(\cos 2 x-1)}{-2 \sin 3 x \sin 2 x} \\
& =\frac{-(1-\cos 2 x)}{-\sin 2 x} \\
& =\frac{1-\cos 2 x}{\sin 2 x} \\
& =\frac{2 \sin ^2 x}{2 \sin x \cos x} \\
& =\frac{\sin x}{\cos x}\\
& =\tan x=\text { R.H.S. } \\
\end{aligned}$
View full question & answer→Question 83 Marks
$\tan 20^{\circ} \tan 80^{\circ} \cot 50^{\circ}=\sqrt{3}$
Answer\begin{aligned}
& \text { L.H.S. }=\tan 20^{\circ} \tan 80^{\circ} \cot 50^{\circ} \\
& =\tan 20^{\circ} \tan 80^{\circ} \cot \left(90^{\circ}-40^{\circ}\right) \\
& =\tan 20^{\circ} \tan 80^{\circ} \tan 40^{\circ} \\
& =\tan 20^{\circ} \tan \left(60^{\circ}+20^{\circ}\right) \tan \left(60^{\circ}-20^{\circ}\right) \\
& =\tan 20^{\circ}\left(\frac{\tan 60^{\circ}+\tan 20^{\circ}}{1-\tan 60^{\circ} \tan 20^{\circ}}\right)\left(\frac{\tan 60^{\circ}-\tan 20^{\circ}}{1+\tan 60^{\circ} \tan 20^{\circ}}\right) \\
& =\tan 20^{\circ}\left(\frac{\sqrt{3}+\tan 20^{\circ}}{1-\sqrt{3} \tan 20^{\circ}}\right)\left(\frac{\sqrt{3}-\tan 20^{\circ}}{1+\sqrt{3} \tan 20^{\circ}}\right) \\
& =\tan 20^{\circ}\left[\frac{(\sqrt{3})^2-\tan { }^2 20^{\circ}}{1^2-\left(\sqrt{3} \tan 20^{\circ}\right)^2}\right] \\
& =\tan 20^{\circ}\left(\frac{3-\tan ^2 20^{\circ}}{1-3 \tan ^2 20^{\circ}}\right) \\
& =\frac{3 \tan 20^{\circ}-\tan ^3 20^{\circ}}{1-3 \tan ^2 20^{\circ}} \\
& =\tan 3\left(20^{\circ}\right) \\
& =\tan 60^{\circ} \\
& =\sqrt{3} \\
& =\text { R. H.S. }
\end{aligned}
View full question & answer→Question 93 Marks
Show that $\sqrt{\frac{1+\sin 2 A}{1-\sin 2 A}}=\tan \left(\frac{\pi}{4}+A\right)$
View full question & answer→Question 103 Marks
Prove that following : $\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}=\frac{1}{16}$
AnswerL.H.S.
$
\begin{aligned}
& =\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ} \\
& =\cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \frac{1}{2} \cdot \cos 80^{\circ} \\
& =\frac{1}{2}\left[\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}\right] \\
& =\frac{1}{4}\left[\cos \left(20^{\circ}+40^{\circ}\right)+\cos \left(20^{\circ}-40^{\circ}\right)\right] \cos 80^{\circ} \\
& \left.=\frac{1}{4}\left[\cos 60^{\circ}\right)+\cos \left(-20^{\circ}\right)\right] \cos 80^{\circ} \\
& =\frac{1}{4}\left[\frac{1}{2} \cos 80^{\circ}+\cos 20^{\circ} \cos 80^{\circ}\right] \\
& =\frac{1}{4}\left[\frac{1}{2} \cos 80^{\circ}+\cos 20^{\circ} \cos 80^{\circ}\right] \\
& =\frac{1}{8} \cos 80^{\circ}+\frac{1}{4} \cdot \frac{1}{2} \cdot 2 \cos 20^{\circ} \cos 80^{\circ} \\
& =\frac{1}{8} \cos 80^{\circ}+\frac{1}{8}[\cos (20+80)+\cos (20-80)] \\
& =\frac{1}{8} \cos 80^{\circ}+\frac{1}{8}\left[\cos 100^{\circ}+\cos (-60)^{\circ}\right] \\
& =\frac{1}{8}\left[\cos 80^{\circ}+\left[\cos 180^{\circ}-80^{\circ}\right)\right]+\frac{1}{8} \times \frac{1}{2} \\
& =\frac{1}{8}\left[\cos 80^{\circ}-\cos 80^{\circ}\right]+\frac{1}{16}=\frac{1}{16} \\
& =\text { R. H.S. }
\end{aligned}
$
View full question & answer→Question 113 Marks
Prove that following : $\frac{\cos 3 x \sin 9 x-\sin x \cos 5 x}{\cos x \cos 5 x-\sin 3 x \sin 9 x}=\tan 8 x$
View full question & answer→Question 123 Marks
Prove that : tan 20° tan 40° tan 60° tan 80° = 3
Question 133 Marks
Show that : $\sec 840^{\circ} \cdot \cot \left(-945^{\circ}\right)+\sin 600^{\circ} \cdot \tan \left(-690^{\circ}\right) \ =\frac{3}{2}$
View full question & answer→Question 143 Marks
Find the value of $\tan \frac{13 \pi}{12}$
View full question & answer→Question 153 Marks
In $\triangle A B C$ prove that
$\tan \frac{A}{2} \tan \frac{B}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{C}{2} \tan \frac{A}{2}=1$
AnswerIn $\triangle A B C, \mathrm{~A}+\mathrm{B}+\mathrm{C}=\pi$
$
\begin{aligned}
\therefore & \mathrm{A}+\mathrm{B}=\pi-\mathrm{C} \therefore \frac{A+B}{2}=\frac{\pi-C}{2}=\frac{\pi}{2}-\frac{C}{2} \\
& \tan \left(\frac{\mathrm{A}}{2}+\frac{B}{2}\right)=\tan \left(\frac{\pi}{2}-\frac{\mathrm{C}}{2}\right) \\
\therefore & \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}=\cot \frac{C}{2} \\
\therefore & \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}} \\
\therefore & {\left[\tan \frac{\mathrm{A}}{2}+\tan \frac{B}{2}\right] \tan \frac{\mathrm{C}}{2}=1-\tan \frac{\mathrm{A}}{2} \tan \frac{B}{2} } \\
\therefore & \tan \frac{\mathrm{A}}{2} \tan \frac{C}{2}+\tan \frac{B}{2} \tan \frac{C}{2}=1-\tan \frac{\mathrm{A}}{2} \tan \frac{B}{2} \\
\therefore & \tan \frac{\mathrm{A}}{2} \tan \frac{C}{2}+\tan \frac{B}{2} \tan \frac{C}{2}+\tan \frac{\mathrm{A}}{2} \tan \frac{B}{2}=1
\end{aligned}
$
View full question & answer→Question 163 Marks
In $\triangle A B C$ prove that
$\sin ^2 \mathrm{~A}+\sin ^2 \mathrm{~B}-\sin ^2 \mathrm{C}=2 \sin \mathrm{A} \sin \mathrm{B} \sin \mathrm{C} $
Answer$\begin{aligned} & \text { L.H.S. }=\sin ^2 \mathrm{~A}+\sin ^2 \mathrm{~B}-\sin ^2 \mathrm{C} \\ & =\frac{1-\cos 2 A}{2}+\frac{1-\cos 2 B}{2}-\sin ^2 C \\ & =\frac{1}{2}[2-\cos 2 \mathrm{~A}-\cos 2 \mathrm{~B}]-\sin ^2 \mathrm{C} \\ & =1-\frac{1}{2}[\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}]-\sin ^2 \mathrm{C} \\ & =1-\frac{1}{2} \cdot 2 \cos \left(\frac{2 A+2 B}{2}\right) \cos \left(\frac{2 A-2 B}{2}\right)-\sin ^2 C \\ & =1-\sin ^2 C-\cos (A+B)+\cos (A-B) \\ & =\cos { }^2 C-\cos [\pi-C] \cos (A-B) \\ & =\cos { }^2 C+\cos C \cos (A-B) \\ & =\cos C[\cos C+\cos (A-B)] \\ & =\cos C[\cos [\pi-(A+B)]+\cos (A-B)] \\ & =\cos C[-\cos (A+B)+\cos (A-B)] \\ & =\cos C[\cos (A-B)-\cos (A+B)]\end{aligned}$
$\begin{aligned} & =\cos C \cdot 2 \sin \left(\frac{A-B+A+B}{2}\right) \sin \left(\frac{A+B-A+B}{2}\right) \\ & =2 \cos C \sin A \sin B \\ & =2 \sin A \sin B \cos C \\ & =\text { R. H.S. }\end{aligned}$
View full question & answer→Question 173 Marks
In $\triangle A B C$ prove that
$\cos A+\cos B+\cos C=1+4 \sin \frac{\mathrm{A}}{2} \sin \frac{\mathrm{B}}{2} \sin \frac{\mathrm{C}}{2}$
AnswerL.H.S $=\cos A+\cos B+\cos C$
$
\begin{aligned}
& =2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)+1-2 \sin ^2 \frac{\mathrm{C}}{2} \\
& =2 \cos \left(\frac{\pi}{2}-\frac{C}{2}\right) \cos \left(\frac{A-B}{2}\right)+1-2 \sin ^2 \frac{\mathrm{C}}{2}
\end{aligned}
$
$\begin{aligned} & =1+2 \sin \frac{C}{2} \cos \left(\frac{A-B}{2}\right)-2 \sin ^2 \frac{C}{2} \\ & =1+2 \sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\sin \frac{C}{2}\right] \\ & =1+2 \sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\sin \left(\frac{\pi}{2}-\frac{A+B}{2}\right)\right] \\ & =1+2 \sin \frac{C}{2}\left[\cos \left(\frac{A-B}{2}\right)-\cos \left(\frac{A+B}{2}\right)\right] \\ & =1+2 \sin \frac{C}{2} \cdot 2 \sin \left(\frac{A-B+A+B}{4}\right) \sin \left(\frac{A+B-A+B}{4}\right) \\ & =1+4 \sin \frac{C}{2} \sin \frac{A}{2} \sin \frac{B}{2} \\ & =1+4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} \\ & =\text { R.H.S. }\end{aligned}$
View full question & answer→Question 183 Marks
Find $\sin \frac{\pi}{10}$
View full question & answer→Question 193 Marks
Prove that $\cos ^2 x+\cos ^2\left(x+\frac{\pi}{3}\right)+\cos ^2\left(x-\frac{\pi}{3}\right)=\frac{3}{2}$
View full question & answer→Question 203 Marks
$\cot \frac{A}{2}+\cot \frac{B}{2}+\cot \frac{C}{2}=\cot \frac{A}{2} \cot \frac{B}{2} \cot \frac{C}{2}$
Answer$\begin{aligned}
& \text { In } \triangle A B C \\
& A+B+C=\pi \\
& \therefore A+B=\pi-C \\
& \therefore \quad \tan \left(\frac{A+B}{2}\right)=\tan \left(\frac{\pi-C}{2}\right) \\
& \therefore \quad \tan \left(\frac{A}{2}+\frac{B}{2}\right)=\tan \left(\frac{\pi}{2}-\frac{C}{2}\right) \\
& \therefore \quad \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \cdot \tan \frac{B}{2}}=\cot \frac{C}{2}
\end{aligned}$
$\begin{aligned}
& \therefore \quad \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \cdot \tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}} \\
& \therefore \quad \tan \frac{C}{2} \cdot\left(\tan \frac{A}{2}+\tan \frac{B}{2}\right)=1-\tan \frac{A}{2} \cdot \tan \frac{B}{2} \\
& \therefore \quad \tan \frac{B}{2} \cdot \tan \frac{C}{2}+\tan \frac{A}{2} \cdot \tan \frac{C}{2}+\tan \frac{A}{2} \cdot \tan \frac{B}{2}=1
\end{aligned}$
Dividing throughout by $\tan \frac{\mathrm{A}}{2} \cdot \tan \frac{\mathrm{B}}{2} \cdot \tan \frac{\mathrm{C}}{2}$, we get
$\begin{aligned}
& \frac{1}{\tan \frac{\mathrm{A}}{2}}+\frac{1}{\tan \frac{\mathrm{B}}{2}}+\frac{1}{\tan \frac{C}{2}}=\frac{1}{\tan \frac{\mathrm{A}}{2} \cdot \tan \frac{\mathrm{B}}{2} \cdot \tan \frac{\mathrm{C}}{2}} \\
& \therefore \quad \cot \frac{\mathrm{A}}{2}+\cot \frac{\mathrm{B}}{2}+\cot \frac{\mathrm{C}}{2}=\cot \frac{\mathrm{A}}{2} \cdot \cot \frac{\mathrm{B}}{2} \cdot \cot \frac{\mathrm{C}}{2}
\end{aligned}$
View full question & answer→Question 213 Marks
$\tan \frac{\mathbf{A}}{2} \tan \frac{\mathbf{B}}{2} \tan \frac{\mathbf{B}}{2} \tan \frac{\mathbf{C}}{2} \tan \frac{\mathbf{C}}{2} \tan \frac{\mathbf{A}}{2}=1$
Answer$\begin{aligned}
& \text { In } \triangle \mathrm{ABC}_{\text {}} \\
& A+B+C=\pi \\
& \therefore A+B=\pi-C \\
& \therefore \quad \tan \left(\frac{A+B}{2}\right)=\tan \left(\frac{\pi-C}{2}\right) \\
& \therefore \quad \tan \left(\frac{A}{2}+\frac{B}{2}\right)=\tan \left(\frac{\pi}{2}-\frac{C}{2}\right) \\
& \therefore \quad \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \cdot \tan \frac{B}{2}}=\cot \frac{C}{2} \\
& \therefore \quad \frac{\tan \frac{A}{2}+\tan \frac{B}{2}}{1-\tan \frac{A}{2} \cdot \tan \frac{B}{2}}=\frac{1}{\tan \frac{C}{2}} \\
& \therefore \quad \tan \frac{C}{2} \cdot\left(\tan \frac{\mathrm{A}}{2}+\tan \frac{\mathrm{B}}{2}\right)=1-\tan \frac{\mathrm{A}}{2} \cdot \tan \frac{\mathrm{B}}{2} \\
& \therefore \quad \tan \frac{A}{2} \cdot \tan \frac{B}{2}+\tan \frac{B}{2} \cdot \tan \frac{C}{2}+\tan \frac{C}{2} \cdot \tan \frac{A}{2}=1 \\
\end{aligned}$
View full question & answer→Question 223 Marks
Prove the following : $\cos 20^{\circ} \cos 40^{\circ} \cos 60^{\circ} \cos 80^{\circ}=1 / 16$
Answer$\begin{aligned}
& \text { L.H.S. }=\cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \cos 60^{\circ} \cdot \cos 80^{\circ} \\
& =\cos 20^{\circ} \cdot \cos 40^{\circ} \cdot \frac{1}{2} \cdot \cos 80^{\circ} \\
& =\frac{1}{2 \times 2}\left(2 \cos 40^{\circ} \cdot \cos 20^{\circ}\right) \cdot \cos 80^{\circ} \\
& =\frac{1}{4}\left[\cos \left(40^{\circ}+20^{\circ}\right)+\cos \left(40^{\circ}-20^{\circ}\right)\right] \cdot \cos 80^{\circ} \\
& =\frac{1}{4}\left(\cos 60^{\circ}+\cos 20^{\circ}\right) \cos 80^{\circ} \\
& =\frac{1}{4} \cos 60^{\circ} \cdot \cos 80^{\circ}+\frac{1}{4} \cos 20^{\circ} \cdot \cos 80^{\circ} \\
& =\frac{1}{4}\left(\frac{1}{2}\right) \cos 80^{\circ}+\frac{1}{2 \times 4}\left(2 \cos 80^{\circ} \cos 20^{\circ}\right) \\
& =\frac{1}{8} \cos 80^{\circ}+\frac{1}{8}\left[\cos \left(80^{\circ}+20^{\circ}\right)+\cos \left(80^{\circ}-20^{\circ}\right)\right] \\
& =\frac{1}{8} \cos 80^{\circ}+\frac{1}{8}\left(\cos 100^{\circ}+\cos 60^{\circ}\right) \\
& =\frac{1}{8} \cos 80^{\circ}+\frac{1}{8} \cos 100^{\circ}+\frac{1}{8} \cos 60^{\circ} \\
& =\frac{1}{8} \cos 80^{\circ}=\frac{1}{8} \cos \left(180^{\circ}-80^{\circ}\right)+\frac{1}{8} \times \frac{1}{2} \\
& =\frac{1}{8} \cos 80^{\circ}-\frac{1}{8} \cos 80^{\circ}+\frac{1}{16} \ldots[\because \cos (180-\theta)=-\cos \theta] \\
& =\frac{1}{16}=R \cdot H \cdot S
\end{aligned}$
View full question & answer→Question 233 Marks
Prove the following : $\frac{\sin x-\sin 3 x+\sin 5 x-\sin 7 x}{\cos x-\cos 3 x-\cos 5 x+\cos 7 x}=\cot 2 x$
Answer$\begin{aligned}
& \text { L.H.S. }=\frac{\sin x-\sin 3 x+\sin 5 x-\sin 7 x}{\cos x-\cos 3 x-\cos 5 x+\cos 7 x} \\
& =\frac{(\sin 5 x+\sin x)-(\sin 7 x+\sin 3 x)}{(\cos x-\cos 5 x)-(\cos 3 x-\cos 7 x)} \\
& =\frac{2 \sin \left(\frac{5 x+x}{2}\right) \cdot \cos \left(\frac{5 x-x}{2}\right)-2 \sin \left(\frac{7 x+3 x}{2}\right) \cdot \cos \left(\frac{7 x-3 x}{2}\right)}{2 \sin \left(\frac{x+5 x}{2}\right) \cdot \sin \left(\frac{5 x-x}{2}\right)-2 \sin \left(\frac{3 x+7 x}{2}\right) \cdot \sin \left(\frac{7 x-3 x}{2}\right)} \\
& =\frac{2 \sin 3 x \cdot \cos 2 x-2 \sin 5 x \cdot \cos 2 x}{2 \sin 3 x \cdot \sin 2 x-2 \sin 5 x \cdot \sin 2 x} \\
& =\frac{2 \cos 2 x(\sin 3 x-\sin 5 x)}{2 \sin 2 x(\sin 3 x-\sin 5 x)} \\
& =\frac{\cos 2 x}{\sin 2 x}=\cot 2 x=\text { R.H.S.S. }
\end{aligned}$
View full question & answer→Question 243 Marks
Prove the following : $4 \cos x \cos \left(\frac{\pi}{3}+x\right) \cos \left(\frac{\pi}{3}-x\right)=\cos 3 x$
Answer$\begin{aligned}
& \text { L.H.S. }=4 \cos x \cdot \cos \left(\frac{\pi}{3}+x\right) \cdot \cos \left(\frac{\pi}{3}-x\right) \\
& =4 \cos x\left(\cos \frac{\pi}{3} \cos x-\sin \frac{\pi}{3} \sin x\right)
\end{aligned}$
( $\left.\cos \frac{\pi}{3} \cos x+\sin \frac{\pi}{3} \sin x\right)$
$\begin{aligned}
& =4 \cos x\left(\frac{1}{2} \cos x-\frac{\sqrt{3}}{2} \sin x\right)\left(\frac{1}{2} \cos x+\frac{\sqrt{3}}{2} \sin x\right) \\
& =4 \cos x\left[\left(\frac{1}{2} \cos x\right)^2-\left(\frac{\sqrt{3}}{2} \sin x\right)^2\right] \\
& \quad=4 \cos x\left(\frac{1}{4} \cos ^2 x-\frac{3}{4} \sin ^2 x\right) \\
& =\cos ^3 x-3 \cos x \cdot \sin ^2 x \\
& =\cos ^3 x-3 \cos x\left(1-\cos ^2 x\right) \\
& =\cos ^3 x-3 \cos x+3 \cos ^3 x \\
& =4 \cos ^3 x-3 \cos x \\
& =\cos 3 x=\text { R.H.S. }
\end{aligned}$
I note: The question has been modified.
View full question & answer→Question 253 Marks
Prove the following : $=\cos ^2 x+\cos ^2\left(x+120^{\circ}\right)+\cos ^2\left(x-120^{\circ}\right)=\frac{3}{2}$
Answer$\begin{aligned}
& \text { L.H.S }=\cos ^2 x+\cos ^2\left(x+120^{\circ}\right)+\cos ^2\left(x-120^{\circ}\right)= \\
& =\frac{1+\cos 2 x}{2}+\frac{1+\cos 2\left(x+120^{\circ}\right)}{2} \\
& +\frac{1+\cos 2\left(x-120^{\circ}\right)}{2} \\
& \cdots\left[\because \cos ^2 \theta=\frac{1+\cos 2 \theta}{2}\right] \\
& \frac{3}{2}+\frac{1}{2}\left[\cos 2 x+\cos \left(2 x+240^{\circ}\right)+\cos \left(2 x 240^{\circ}\right)\right] \\
& =\frac{3}{2}+\frac{1}{2}\left(\cos 2 x+\cos 2 x \cos 240^{\circ}-\sin 2 x \sin 240^{\circ}+\cos 2 x \cos \right. \\
& \left.240^{\circ}+\sin 2 x \sin 240^{\circ}\right) \\
& =\frac{3}{2}+\frac{1}{2}\left(\cos 2 x+2 \cos 2 x \cos 240^{\circ}\right) \\
& =\frac{3}{2}+\frac{1}{2}\left[\cos 2 x+2 \cos 2 x \cos \left(180^{\circ}+60^{\circ}\right)\right] \\
& =\frac{3}{2}+\frac{1}{2}[\cos 2 x+2 \cos 2 x(-\cos 600)] \\
& =\frac{3}{2}+\frac{1}{2}\left[\cos 2 x-2 \cos 2 x\left(\frac{1}{2}\right)\right] \\
& =\frac{3}{2}+\frac{1}{2}(\cos 2 x-\cos 2 x) \\
& =\frac{3}{2}+\frac{1}{2}(0) \\
& =\frac{3}{2}=\text { R.H.S. } \\
\end{aligned}$
View full question & answer→Question 263 Marks
Prove the following : $\frac{2 \cos 4 x+1}{2 \cos x+1}=(2 \cos x-1)(2 \cos 2 x-1)$
Answer$\begin{aligned}
& \text { L.H.S. }=\frac{2 \cos 4 x+1}{2 \cos x+1} \\
& =\frac{2\left[2 \cos ^2(2 x)-1\right]+1}{2 \cos x+1} \\
& \ldots\left[\because \cos 2 \theta=2 \cos ^2 \theta-1\right] \\
& =\frac{4 \cos ^2 2 x-2+1}{2 \cos x+1} \\
& =\frac{(2 \cos 2 x)^2-(1)^2}{2 \cos x+1} \\
& =\frac{(2 \cos 2 x+1)(2 \cos 2 x-1)}{2 \cos x+1} \\
& =\frac{\left[2\left(2 \cos ^2 x-1\right)+1\right](2 \cos 2 x-1)}{2 \cos x+1} \\
& =\frac{\left(4 \cos ^2 x-2+1\right)(2 \cos 2 x-1)}{2 \cos x+1} \\
& =\frac{\left[(2 \cos x)^2-(1)^2\right](2 \cos 2 x-1)}{2 \cos x+1} \\
& =\frac{(2 \cos x+1)(2 \cos x-1)(2 \cos 2 x-1)}{2 \cos x+1} \\
& =(2 \cos x-1)(2 \cos 2 x-1) \\
& =\mathrm{R} \cdot \mathrm{H} \cdot \mathrm{S} \text {. } \\
\end{aligned}$
View full question & answer→Question 273 Marks
Prove the following : $=\frac{\sin ^2\left(-160^{\circ}\right)}{\sin ^2 70^{\circ}}+\frac{\sin \left(180^{\circ}-\theta\right)}{\sin \theta}=\sec ^2 20^{\circ}$
Answer$\begin{aligned}
& \text { L.H.S. }=\frac{\sin ^2\left(-160^{\circ}\right)}{\sin ^2 70^{\circ}}+\frac{\sin \left(180^{\circ}-\theta\right)}{\sin \theta} \\
&=\frac{\left(-\sin 160^{\circ}\right)^2}{\sin ^2 70^{\circ}}+\frac{\sin \theta}{\sin \theta} \\
&=\frac{\sin ^2 160^{\circ}}{\sin ^2 70^{\circ}}+1\\
&=1+\frac{\left(\sin 160^{\circ}\right)^2}{\left(\sin 70^{\circ}\right)^2} \\
&= 1+\frac{\left[\sin ^{\circ}\left(180^{\circ}-20^{\circ}\right)\right]^2}{\left[\sin ^{\circ}\left(90^{\circ}-20^{\circ}\right)\right]^2} \\
&=1+\frac{\sin ^2 20^{\circ}}{\cos ^2 20^{\circ}} \\
&= 1+\tan ^2 20^{\circ} \\
&= \sec ^2 2^{\circ}=\text { R.H.S. }
\end{aligned}$
View full question & answer→Question 283 Marks
Prove the following : $\cos 7^{\circ} \cos 14^{\circ} \cos 28^{\circ} \cos 56^{\circ}=\frac{\sin 68^{\circ}}{16 \cos 83^{\circ}}$
Answer$\begin{aligned}
& \text { L.H.S. }=\cos 7^{\circ} \cos 14^{\circ} \cos 28^{\circ} \cos 56^{\circ} \\
& =\frac{1}{2 \sin 7^{\circ}}\left(2 \sin 7^{\circ} \cos 7^{\circ}\right) \cos 14^{\circ} \cos 28^{\circ} \cos 56^{\circ} \\
& =\frac{1}{2 \sin 7^{\circ}}\left(\sin 14^{\circ} \cos 14^{\circ} \cos 28^{\circ} \cos 56^{\circ}\right) \ldots[\because 2 \sin \theta \cos \theta=\sin 2 \theta] \\
& =\frac{1}{2\left(2 \sin 7^{\circ}\right)}\left(2 \sin 14^{\circ} \cos 14^{\circ}\right) \cos 28^{\circ} \cos 56^{\circ} \\
& =\frac{1}{4 \sin 7^0}\left(\sin 28^{\circ} \cos 28^{\circ} \cos 56^{\circ}\right) \\
& =\frac{1}{2\left(4 \sin 7^{\circ}\right)}\left(2 \sin 28^{\circ} \cos 28^{\circ}\right) \cos 56^{\circ} \\
& =\frac{1}{8 \sin 7^{\circ}}\left(\sin 56^{\circ} \cos 56^{\circ}\right) \\
& =\frac{1}{2\left(8 \sin 7^{\circ}\right)}\left(2 \sin 56^{\circ} \cos 56^{\circ}\right) \\
& =\frac{1}{16 \sin 7^{\circ}}\left(\sin 112^{\circ}\right) \\
& =\frac{\sin \left(180^{\circ}-68^{\circ}\right)}{16 \sin \left(90^{\circ}-83^{\circ}\right)} \\
& =\frac{\sin 68^{\circ}}{16 \cos 83^{\circ}} \\
& =R \cdot-5.5
\end{aligned}$
View full question & answer→Question 293 Marks
Prove the following : $\frac{\sin ^3(\pi+x) \sec ^2(\pi-x) \tan (2 \pi-x)}{\cos ^2\left(\frac{\pi}{2}+x\right) \sin (\pi-x) \operatorname{cosec}^2(-x)}=\tan ^3 x$
Answer$
\begin{aligned}
& \text { L.H.S. }=\frac{\sin ^3(\pi+x) \sec ^2(\pi-x) \tan (2 \pi-x)}{\cos ^2\left(\frac{\pi}{2}+x\right) \sin (\pi-x) \operatorname{cosec}^2(-x)} \\
& =\frac{[\sin (\pi+x)]^3[\sec (\pi-x)]^2 \tan (2 \pi-x)}{\left[\cos \left(\frac{\pi}{2}+x\right)\right]^2 \sin (\pi-x) \cdot(-\operatorname{cosec} x)^2} \\
& =\frac{(-\sin x)^3(-\sec x)^2(-\tan x)}{(-\sin x)^2 \cdot \sin x \cdot \operatorname{cosec}^2 x} \\
& =\frac{\left(-\sin ^3 x\right) \cdot \sec ^2 x \cdot(-\tan x)}{\sin ^2 x \cdot \sin x \cdot \frac{1}{\sin ^2 x}} \\
& =\frac{\sin ^3 x \cdot \sec ^2 x \cdot \tan x}{\sin x} \\
& =\sin ^2 x \cdot \frac{1}{\cos ^2 x} \cdot \tan x \\
& =\tan ^2 x \cdot \tan x \\
& =\tan ^3 x \\
& =\mathbf{R.H.S}
\end{aligned}
$
View full question & answer→Question 303 Marks
If $\tan A=\frac{5}{6}, \tan B=\frac{1}{11}$ prove that $A+B=\frac{\pi}{4}$
AnswerGiven $\tan A=\frac{5}{6}, \tan B=\frac{1}{11}$
$\begin{aligned}
& \tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \tan B}=\frac{\frac{5}{6}+\frac{1}{11}}{1-\left(\frac{5}{6}\right)\left(\frac{1}{11}\right)} \\
&=\frac{\left(\frac{61}{66}\right)}{\left(\frac{61}{66}\right)}=1 \\
& \therefore \tan (A+B)=\tan \frac{\pi}{4} \\
& \therefore A+B=\frac{\pi}{4}
\end{aligned}$
View full question & answer→Question 313 Marks
Prove the following : $\tan 10^{\circ}+\tan 35^{\circ}+\tan 10^{\circ} \cdot \tan 35^{\circ}=1$
Answer$\begin{aligned}
\text { L.H.S. } & =\frac{\cot A \cot 4 A+1}{\cot A \cot 4 A-1} \\
& =\frac{\frac{\cos A}{\sin A} \cdot \frac{\cos 4 A}{\sin 4 A}+1}{\frac{\cos A}{\sin A} \cdot \frac{\cos 4 A}{\sin 4 A}-1} \\
& =\frac{\frac{\cos A \cos 4 A+\sin A \sin 4 A}{\sin A \sin 4 A}}{\frac{\cos A \cos 4 A-\sin A \sin 4 A}{\sin A \sin 4 A}} \\
& =\frac{\cos 4 A \cos A+\sin 4 A \sin A}{\cos 4 A \cos A-\sin 4 A \sin A} \\
& =\frac{\cos (4 A-A)}{\cos (4 A+A)} \\
& =\frac{\cos 3 A}{\cos 5 A}=\text { R.H.S. }
\end{aligned}$
View full question & answer→Question 323 Marks
Prove the following : $\frac{\cos 27^{\circ}+\sin 27^{\circ}}{\cos 27^{\circ}-\sin 27^{\circ}}=\tan 72^{\circ}$
Answer$\frac{\cos 27^{\circ}+\sin 27^{\circ}}{\cos 27^{\circ}-\sin 27^{\circ}}$
Dividing numerator and $\cos 27^{\circ}$, we get denominator by $\cos$ $27^{\circ}$, we get
$
\begin{aligned}
\text { L.H.S. } & =\frac{1+\frac{\sin 27^{\circ}}{\cos 27^{\circ}}}{1-\frac{\sin 27^{\circ}}{\cos 27^{\circ}}} \\
& =\frac{1+\tan 27^{\circ}}{1-\tan 27^{\circ}} \\
& =\frac{\tan 45^{\circ}+\tan 27^{\circ}}{1-\tan 45^{\circ} \tan 27^{\circ}} \quad \ldots\left[\because \tan 45^{\circ}=1\right] \\
& =\tan \left(45^{\circ}+27^{\circ}\right) \\
& =\tan 72^{\circ}=\text { R.H.S. } \\
=\tan \left(45^{\circ}\right. & \left.+27^{\circ}\right) \\
=\tan 72^{\circ} & =\text { R.H.S }
\end{aligned}
$
View full question & answer→Question 333 Marks
Prove the following : $\tan 50^{\circ}=\tan 40^{\circ}+2 \tan 10^{\circ}$
Answer\begin{aligned}
& \text { Since, } 50^{\circ}=10^{\circ}+40^{\circ} \\
& \therefore \tan 50^{\circ}=\tan \left(10^{\circ}+40^{\circ}\right) \\
& \therefore \frac{\tan 10^{\circ}+\tan 40^{\circ}}{1-\tan 10^{\circ} \tan 40^{\circ}} \\
& \therefore \tan 50^{\circ}\left(1-\tan 10^{\circ} \tan 40^{\circ}\right)=\tan 10^{\circ}+\tan 40^{\circ} \\
& \therefore \tan 50^{\circ}-\tan 10^{\circ} \tan 40^{\circ} \tan 50^{\circ}=\tan 10^{\circ}+\tan 40^{\circ} \\
& \therefore \tan 50^{\circ}-\tan 10^{\circ} \tan 40^{\circ} \tan \left(90^{\circ}-40^{\circ}\right)=\tan 10^{\circ}+\tan \\
& 40^{\circ} \\
& \therefore \tan 50^{\circ}-\tan 10^{\circ} \tan 40^{\circ} \cot 40^{\circ} \\
& =\tan 10^{\circ}+\tan 40^{\circ} \ldots\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right] \\
& \therefore \tan 50^{\circ}-\tan 10^{\circ} \tan 40^{\circ} \cdot \frac{1}{\tan 40^{\circ}}=\tan 10^{\circ}+\tan 40^{\circ} \\
& \therefore \tan 50^{\circ}-\tan 10^{\circ} \cdot 1=\tan 10^{\circ}+\tan 40^{\circ} \\
& \therefore \tan 50^{\circ}=\tan 40^{\circ}+2 \tan 10^{\circ}
\end{aligned}
View full question & answer→Question 343 Marks
Prove the following : $\frac{\tan 5 A-\tan 3 A}{\tan 5 A+\tan 3 A}=\frac{\sin 2 A}{\sin 8 A}$
Answer$
\begin{aligned}
\text { L.H.S. } & =\frac{\tan 5 \mathrm{~A}-\tan 3 \mathrm{~A}}{\tan 5 \mathrm{~A}+\tan 3 \mathrm{~A}} \\
& =\frac{\frac{\sin 5 \mathrm{~A}}{\cos 5 \mathrm{~A}}-\frac{\sin 3 \mathrm{~A}}{\cos 3 \mathrm{~A}}}{\frac{\sin 5 \mathrm{~A}}{\cos 5 \mathrm{~A}}+\frac{\sin 3 \mathrm{~A}}{\cos 3 \mathrm{~A}}}
\end{aligned}
$
$
\begin{aligned}
& =\frac{\left(\frac{\sin 5 A \cos 3 A-\sin 3 A \cos 5 A}{\cos 5 A \cos 3 A}\right)}{\left(\frac{\sin 5 A \cos 3 A+\sin 3 A \cos 5 A}{\cos 5 A \cos 3 A}\right)} \\
& =\frac{\sin 5 A \cos 3 A-\cos 5 A \sin 3 A}{\sin 5 A \cos 3 A+\cos 5 A \sin 3 A} \\
& =\frac{\sin (5 A-3 A)}{\sin (5 A+3 A)} \\
& =\frac{\sin 2 A}{\sin 8 A} \\
& =\text { R.H.S. }
\end{aligned}
$
View full question & answer→