Question : If $\sec 2 \theta=\operatorname{cosec}(\theta-36°)$, where $2 \theta$ is an acute angle, find the value of $\theta$.
Option 1: 32°
Option 2: 46°
Option 3: 20°
Option 4: 42°
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Correct Answer: 42°
Solution : Given, $\sec 2 \theta=\operatorname{cosec}\left(\theta-36° \right)$ We know, $ \operatorname{cosec}\theta = \operatorname{sec}(90°-\theta)$ ⇒ $\sec2\theta = \sec\{90°-(\theta-36°)\}$ ⇒ $\sec2\theta=\sec(90°-\theta+36°)$ ⇒ $2\theta=126°-\theta$ ⇒ $3\theta=126°$ ⇒ $\theta=\frac{126°}{3}=42°$ Hence, the correct answer is 42°.
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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: $\operatorname{cosec} \theta$
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 : $\left(\frac{\tan ^3 \theta}{\sec ^2 \theta}+\frac{\cot ^3 \theta}{\operatorname{cosec}^2 \theta}+2 \sin \theta \cos \theta\right) \div\left(1+\operatorname{cosec}^2 \theta+\tan ^2 \theta\right), 0^{\circ}<\theta<90^{\circ}$, is equal to:
Option 1: $\operatorname{cosec} \theta \sec \theta$
Option 2: $\operatorname{cosec} \theta$
Option 3: $\sin \theta \cos \theta$
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: $1$
Question : $\theta$ is a positive acute angle and $\sin\theta-\cos\theta=0$, then the value of $\sec\theta+\operatorname{cosec}\theta$ is:
Option 1: $2$
Option 2: $\sqrt{3}$
Option 3: $2\sqrt{2}$
Option 4: $3\sqrt{2}$
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