## Trigonometry Questions and Answers Part-8

1. $6\left(\sin ^{6}\theta+\cos^{6}\theta\right)-9\left(\sin ^{4}\theta+\cos^{4}\theta\right)$
is equal to
a) -1
b) 1
c) -3
d) 3

Explanation:

2. $\left(1+\tan\alpha\tan\beta\right)^{2}+\left(\tan\alpha-\tan\beta\right)^{2}$
is equal to
a) $\tan ^{2}\alpha+\tan ^{2}\beta$
b) $\cos ^{2}\alpha\cos ^{2}\beta$
c) $\sec ^{2}\alpha \sec ^{2}\beta$
d) $\tan ^{2}\alpha\tan ^{2}\beta$

Explanation:

3. If $\tan\alpha =\frac{5}{6}\tan\beta=\frac{1}{11}$     then
a) $\alpha+\beta=\pi/6$
b) $\alpha+\beta=\pi/4$
c) $\alpha+\beta=\pi/3$
d) none of these

Explanation:

4. $\sqrt{2+\sqrt{2+2\cos 4\theta}}$     is equal to
a) $\cos \theta$
b) $2\cos \theta$
c) $\cos 2\theta$
d) $2\cos 2\theta$

Explanation:

5. If tan x tan y = a and x + y = $\pi$ /6, then tan x and tan y satisfy the equation
a) $x^{2}-\sqrt{3}\left(1-a\right)x+a=0$
b) $\sqrt{3}x^{2}-\left(1-a\right)x+a \sqrt{3}=0$
c) $x^{2}+\sqrt{3}\left(1+a\right)x-a=0$
d) $\sqrt{3}x^{2}+\left(1+a\right)x-a \sqrt{3}=0$

Explanation:

6. $\frac{\sin 7 x+6\sin 5 x+17\sin 3 x+12\sin x}{\sin 6 x+5\sin 4 x+12\sin 2 x}$
is equal to
a) $\cos x$
b) $2\cos x$
c) $\sin x$
d) $2\sin x$

Explanation:

7. If $\tan \beta=\frac{n\sin \alpha\cos\alpha}{1-n\cos^{2}\alpha}$     then, $\tan\left(\alpha+\beta\right)$   is equal to
a) $\left(n-1\right)\tan \alpha$
b) $\left(n+1\right)\tan \alpha$
c) $\frac{1}{n+1}\tan \alpha$
d) $\frac{-1}{n-1}\tan \alpha$

Explanation:

8. If $\sin\alpha+\cos\alpha=\frac{\sqrt{7}}{2},0<\alpha <\frac{\pi}{6}$
then $\tan=\frac{\alpha}{2}$   is equal to
a) $\sqrt{7}-2$
b) $\left(1/3\right)\left(\sqrt{7}-2\right)$
c) $2-\sqrt{7}$
d) $\left(1/3\right)\left(2-\sqrt{7}\right)$

Explanation:

9. If $\sin\alpha=\frac{336}{625}$     and $450^{\circ}<\alpha < 540^{\circ}$    ,then $\sin\left(\alpha/4\right)$   is equal to
a) $\frac{1}{5\sqrt{2}}$
b) $\frac{7}{25}$
c) $\frac{4}{5}$
d) $\frac{3}{5}$

10. If $\sin \left(\theta+\alpha\right)=a$    and $\sin \left(\theta+\beta\right)=b$    , $\left(0<\alpha,\beta,\theta< \pi/2\right)$     then
$\cos 2\left(\alpha-\beta\right)-4ab\cos \left(\alpha-\beta\right)$       is equal to
a) $1-a^{2}-b^{2}$
b) $1-2a^{2}-2b^{2}$
c) $2+a^{2}+b^{2}$
d) $2-a^{2}-b^{2}$