Question : If $\operatorname{cosec} \theta+\cot \theta=p$, then the value of $\frac{p^2-1}{p^2+1}$ is:
Option 1: $\cos \theta$
Option 2: $\sin \theta$
Option 3: $\cot \theta$
Option 4: $\operatorname{cosec} \theta$
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Correct Answer: $\cos \theta$
Solution : Given, $\operatorname{cosec} \theta+\cot \theta=p$ Squaring both sides, we get, $\operatorname{cosec}^2 \theta+\cot^2 \theta+2\operatorname{cosec} \theta\cot \theta=p^2$ ----------------------(1) Adding 1 on both sides of equation (1), $\operatorname{cosec}^2 \theta+(\cot^2 \theta+1)+2\operatorname{cosec} \theta\cot \theta=p^2+1$ $⇒2\operatorname{cosec}^2 \theta+2\operatorname{cosec} \theta\cot \theta=p^2+1$ ---------------------------(2) Subtracting 1 from both sides of equation (1), $(\operatorname{cosec}^2-1) \theta+\cot^2 \theta+2\operatorname{cosec} \theta\cot \theta=p^2-1$ $⇒2\cot^2 \theta+2\operatorname{cosec} \theta\cot \theta=p^2-1$-------------------------------- (3) Dividing equation (3) by equation (2), $\frac{2\cot^2 \theta+2\operatorname{cosec} \theta\cot \theta}{2\operatorname{cosec}^2 \theta+2\operatorname{cosec} \theta\cot \theta}=\frac{p^2-1}{p^2+1}$ $⇒\frac{2\cot \theta(\cot \theta +\operatorname{cosec} \theta)}{2\operatorname{cosec} \theta(\cot \theta + \operatorname{cosec} \theta)}=\frac{p^2-1}{p^2+1}$ $⇒\frac{\cot \theta}{\operatorname{cosec} \theta}=\frac{p^2-1}{p^2+1}$ $\therefore\frac{p^2-1}{p^2+1}=\cos \theta$ Hence, the correct answer is $\cos \theta$.
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Question : The expression $\frac{\cos ^4 \theta-\sin ^4 \theta+2 \sin ^2 \theta+3}{(\operatorname{cosec} \theta+\cot \theta+1)(\operatorname{cosec} \theta-\cot \theta+1)-2}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : $\frac{1+\cos \theta-\sin ^2 \theta}{\sin \theta(1+\cos \theta)} \times \frac{\sqrt{\sec ^2 \theta+\operatorname{cosec}^2 \theta}}{\tan \theta+\cot \theta}, 0^{\circ}<\theta<90^{\circ}$, is equal to:
Question : The value of $\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$ is:
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:
Question : If $\operatorname{cosec} \theta + \operatorname{cot} \theta = m$, find the value of$\frac{m^2 – 1}{m^2 + 1}$
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