Question : If $\cos \theta=\frac{\sqrt{3}}{2}$, then $\tan ^2 \theta \cos ^2 \theta=?$
Option 1: $\frac{1}{\sqrt{3}}$
Option 2: $\frac{1}{4}$
Option 3: $\frac{1}{2}$
Option 4: $\sqrt{3}$
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Correct Answer: $\frac{1}{4}$
Solution : We have, $\cos \theta = \frac{\sqrt{3}}{2}$ ⇒ $\cos \theta = \cos 30^\circ$ ⇒ $\theta = 30^\circ$ Now, $\tan^2 \theta \cos^2 \theta=\tan^2 30^\circ \cos^2 30^\circ= (\frac{1}{\sqrt3})^2 ×(\frac{\sqrt{3}}{2})^2$ $= \frac{1}{4}$ Hence, the correct answer is $\frac{1}{4}$.
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Question : If $\frac{\sin ^2 \theta}{\cos ^2 \theta-3 \cos \theta+2}=1, \theta$ lies in the first quadrant, then the value of $\frac{\tan ^2 \frac{\theta}{2}+\sin ^2 \frac{\theta}{2}}{\tan \theta+\sin \theta}$ is:
Question : If $2 \cot \theta = 3$, find the value of $\frac{\sqrt{13} \sin \theta – 3 \tan \theta}{3 \tan \theta + \sqrt{13} \cos \theta}$
Question : If $\cos\theta+\sin\theta=\sqrt{2}\cos\theta$, then $\cos\theta-\sin\theta$ is:
Question : If $7 \sin ^2 \theta+4 \cos ^2 \theta=5$ and $\theta$ lies in the first quadrant, then what is the value of $\frac{\sqrt{3} \sec \theta+\tan \theta}{\sqrt{2} \cot \theta-\sqrt{3} \cos \theta}$?
Question : What is $\tan \frac{\theta}{2}$?
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