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) \[\frac{3}{5}\]
Explanation:
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) \[2-a^{2}-b^{2}\]
Explanation:
