Sample QuestionsTrigonometry – II questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
The numerical value of $\tan 20^{\circ} \tan 80^{\circ} \cot 50^{\circ}$ is equal to
- ✓
$\sqrt{ } 3$
- B
$\frac{1}{\sqrt{3}}$
- C
$2 \sqrt{3}$
- D
$\frac{1}{2 \sqrt{3}}$
Answer: A.
View full solution →In $\triangle A B C$ if $\cot A \cot B \cot C>0$, then the triangle is
Answer: A.
View full solution →If $\alpha+\beta+\gamma=\pi$, then the value of $\sin ^2 \alpha+\sin ^2 \beta-\sin ^2 \gamma$ is equal to
- A
$2 \sin \alpha$
- B
$2 \sin \alpha \cos \beta \sin \gamma$
- ✓
$2 \sin \alpha \sin \beta \cos \gamma$
- D
$2 \sin \alpha \sin \beta \sin \gamma$
Answer: C.
View full solution →The value of $\sin \frac{\pi}{14} \sin \frac{3 \pi}{14} \sin \frac{5 \pi}{14} \sin \frac{7 \pi}{14} \sin \frac{9 \pi}{14} \sin \frac{11 \pi}{14} \sin \frac{13 \pi}{14}$ is
- A
$\frac{1}{16}$
- ✓
$\frac{1}{64}$
- C
$\frac{1}{128}$
- D
$\frac{1}{256}$
Answer: B.
View full solution →The value of $\cos A \cos \left(60^{\circ}-A\right) \cos \left(60^{\circ}+A\right)$ is equal to
Answer: C.
View full solution →$\sin 36^{\circ}=\frac{\sqrt{10-2 \sqrt{5}}}{4}$
View full solution →Show that $\\4 \sin \theta \cos ^3 \theta-4 \cos \theta \sin ^3 \theta=\sin 4 \theta$
View full solution →Show that $\cos \left(\frac{\pi}{4}+x\right)+\cos \left(\frac{\pi}{4}-x\right)=\sqrt{2} \cos x$
View full solution →Prove that $\frac{\cos ^3 \theta-\cos 3 \theta}{\cos \theta}+\frac{\sin ^3 \theta+\sin 3 \theta}{\sin \theta}=3$
View full solution →$\tan 3 x \tan 2 x \tan x=\tan 3 x-\tan 2 x-\tan x$
View full solution →$\sqrt{ } 3 \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}=4$
View full solution →$\sin 47^{\circ}+\sin 61^{\circ}-\sin 11^{\circ}-\sin 25^{\circ}=\cos 7^{\circ}$
View full solution →$\tan \frac{\pi}{8}=\sqrt{2}-1$
View full solution →$\sin 36^{\circ}=\frac{\sqrt{10-2 \sqrt{5}}}{4}$
View full solution →$\cos 36^{\circ}=\frac{\sqrt{5}+1}{4}$
View full solution →$\tan 6^{\circ} \tan 42^{\circ} \tan 66^{\circ} \tan 78^{\circ}=1$
View full solution →$\sin \frac{\pi^c}{8}=\frac{1}{2} \sqrt{2-\sqrt{2}}$
View full solution →$\sin 18^{\circ}=\frac{\sqrt{5}-1}{4}$
View full solution →$\sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}=\frac{\sqrt{3}}{8}$
View full solution →$3 \tan ^6 10^{\circ}-27 \tan ^4 10^{\circ}+33 \tan ^2 10^{\circ}=1$
View full solution →Prove The Theorem : For any two angles A and B, $\tan (\mathrm{A}-\mathrm{B})=\frac{\tan A-\tan B}{1+\tan A \tan B}$ (Activity)
View full solution →Prove The Theorem : For any two angles A and B,$\tan (\mathrm{A}+\mathrm{B})=\frac{\tan A+\tan B}{1-\tan A \tan B}$
View full solution →Prove The Theorem : For any two angles A and B,
$\sin (A+B)=\sin A \cos B+\cos A \sin B \quad \text { [verify] }$
View full solution →Theorem : For any two angles A and B,
$\sin (A-B)=\sin A \cos B-\cos A \sin B$
View full solution →Theorem : For any two angles A and B,
$\cos (A+B)=\cos A \cos B-\sin A \sin B$
View full solution →$\frac{\tan ^2 x}{1+\tan ^2 x}+\frac{\cot ^3 x}{1+\cot ^2 x}=\sec x \operatorname{cosec} x-2 \sin x \cos x$
View full solution →In any triangle $\mathrm{ABC}_r \sin \mathrm{A}-\cos \mathrm{B}=\cos \mathrm{C}$. Show that $\angle \mathrm{B}=\frac{\pi}{2}$.
View full solution →If $A+B+C=\frac{3 \pi}{2}$, then $\cos 2 A+\cos 2 B+\cos 2 C=1-4 \sin A$ $\sin B \sin C$
View full solution →$\tan \mathrm{A}+\tan \left(60^{\circ}+\mathrm{A}\right)+\tan \left(120^{\circ}+\mathrm{A}\right)=3 \tan 3 \mathrm{~A}$
View full solution →$\frac{2 \cos 2 \mathrm{~A}+1}{2 \cos 2 \mathrm{~A}-1}=\tan \left(60^{\circ}+\mathrm{A}\right) \tan \left(60^{\circ}-\mathrm{A}\right) $
View full solution →In $\triangle \mathrm{ABC}, \angle \mathrm{C}=\frac{2 \pi}{3}$, then prove that $\cos ^2 \mathrm{~A}+\cos ^2 \mathrm{~B}-\cos \mathrm{A} \cos \mathrm{B}=\frac{3}{4}$.
View full solution →$\tan A+2 \tan 2 A+4 \tan 4 A+8 \cot 8 \mathrm{~A}=\cot A $
View full solution →$3(\sin x-\cos x)^4+6(\sin x+\cos x)^2+4\left(\sin ^6 x+\cos ^6 x\right)=13 $
View full solution →$ \text { cosec } 48^{\circ}+\operatorname{cosec} 96^{\circ}+\operatorname{cosec} 192^{\circ}+\operatorname{cosec} 384^{\circ}=0 $
View full solution →$\cos \frac{2 \pi}{15} \cos \frac{4 \pi}{15} \cos \frac{8 \pi}{15} \cos \frac{16 \pi}{15}=\frac{1}{16}$
View full solution →