Question : The value of $\tan ^2 48^{\circ}-\operatorname{cosec}^2 42^{\circ}+\operatorname{cosec}\left(67^{\circ}+\theta\right)-\sec \left(23^{\circ}-\theta\right)$ is:
Option 1: $-1$
Option 2: $0$
Option 3: $1$
Option 4: $-2$
Correct Answer: $-1$
Solution : Given: $\tan ^2 48^{\circ}-\operatorname{cosec}^2 42^{\circ}+\operatorname{cosec}\left(67^{\circ}+\theta\right)-\sec \left(23^{\circ}-\theta\right)$ = $\tan ^2 48^{\circ}-\sec^2 (90^{\circ}-42^{\circ})+\sec(90^{\circ}-(67^{\circ}+\theta))-\sec \left(23^{\circ}-\theta\right)$ = $\tan ^2 48^{\circ}-\sec^2 48^{\circ}+\sec(23^{\circ}-\theta)-\sec \left(23^{\circ}-\theta\right)$ = $-1$ Hence, the correct answer is $-1$.
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Question : If $\frac{1}{\operatorname{cosec} \theta+1}+\frac{1}{\operatorname{cosec} \theta-1}=2 \sec \theta, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{\tan \theta+2 \sec \theta}{\operatorname{cosec} \theta}$ is:
Option 1: $\frac{4+\sqrt{2}}{2}$
Option 2: $\frac{2+\sqrt{3}}{2}$
Option 3: $\frac{4+\sqrt{3}}{2}$
Option 4: $\frac{2+\sqrt{2}}{2}$
Question : If $\cot \theta=\frac{1}{\sqrt{3}}, 0^{\circ}<\theta<90^{\circ}$, then the value of $\frac{2-\sin ^2 \theta}{1-\cos ^2 \theta}+\left(\operatorname{cosec}^2 \theta-\sec \theta\right)$ is:
Option 1: 0
Option 2: 2
Option 3: 5
Option 4: 1
Question : Find the value of $\left(\tan ^2 \theta+\tan ^4 \theta\right)$.
Option 1: $\cot ^2 \theta-\tan ^2 \theta$
Option 2: $\ {\sec}^4 \theta-\ {\sec}^2 \theta$
Option 3: $\ {\sec}^4 \theta-\ {\sec}^4 \theta$
Option 4: $ \ {\sec}^4 \theta+\ {\sec}^2 \theta$
Question : Solve the following equation. $\sec ^2 \theta\left(\sqrt{1-\sin ^2 \theta}\right)= $ __________.
Option 1: $\tan \theta$
Option 2: $\operatorname{cosec \theta}$
Option 3: $\sec \theta$
Option 4: $1$
Question : The value of $\left(2 \cos ^2 \theta-1\right)\left[\frac{1+\tan \theta}{1-\tan \theta}+\frac{1-\tan \theta}{1+\tan \theta}\right]$ is:
Option 1: $2$
Option 3: $\frac{\sqrt{3}}{2}$
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