Question : Find the value of $\tan 3 \theta$, if $\sec 3 \theta=\operatorname{cosec}\left (4 \theta-15°\right)$.
Option 1: $\frac{1}{\sqrt{3}}$
Option 2: $\sqrt3$
Option 3: $–1$
Option 4: $1$
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Correct Answer: $1$
Solution : Given, $\sec 3 \theta=\operatorname{cosec}\left (4 \theta-15°\right)$ We know $ \operatorname{cosec}\theta = \sec(90°-\theta)$ ⇒ $\sec3\theta = \sec (90°-(4\theta - 15°))$ ⇒ $3\theta = 90°-4\theta + 15°$ ⇒ $7\theta = 105°$ ⇒ $\theta = 15°$ $\therefore$ $\tan 3 \theta=\tan (3\times15)=\tan 45° = 1$ Hence, the correct answer is $1$.
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Question : $\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Option 1: $\operatorname{cosec} \theta \sec \theta$
Option 2: $\operatorname{cosec} \theta$
Option 3: $\sin \theta \cos \theta$
Option 4: $\sec \theta$
Question : The value of $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$ is:
Option 1: $\sec\theta+\tan \theta$
Option 2: $\operatorname{cosec} \theta-\cot \theta$
Option 3: $\operatorname{cosec} \theta+\cot \theta$
Option 4: $\sec\theta-\tan \theta$
Question : The value of $\frac{3\left(\operatorname{cosec}^2 26^{\circ}-\tan ^2 64^{\circ}\right)+\left(\cot ^2 42^{\circ}-\sec ^2 48^{\circ}\right)}{\cot \left(22^{\circ}-\theta\right)-\operatorname{cosec}^2\left(62^{\circ}+\theta\right)-\tan \left(\theta+68^{\circ}\right)+\tan ^2\left(28^{\circ}-\theta\right)}$ is:
Option 1: 3
Option 2: 4
Option 3: –1
Option 4: –2
Question : $\frac{(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)}{(\sec \theta+\tan \theta)(1-\sin \theta)}$ is equal to:
Option 1: $2 \sec \theta$
Option 2: $2 \operatorname{cosec} \theta$
Option 3: $\operatorname{cosec} \theta$
Question : If $\sec\theta-\tan\theta=\frac{1}{\sqrt3}$, then the value of $\sec\theta.\tan\theta$ is:
Option 1: $\frac{2}{3}$
Option 2: $\frac{2}{\sqrt3}$
Option 3: $\frac{4}{\sqrt3}$
Option 4: $\frac{1}{\sqrt3}$
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