Question : If $\operatorname{cosec} \theta + \operatorname{cot} \theta = m$, find the value of$\frac{m^2 – 1}{m^2 + 1}$
Option 1: $1$
Option 2: $0$
Option 3: $\operatorname{cos} \theta$
Option 4: $\operatorname{–sin} \theta$
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Correct Answer: $\operatorname{cos} \theta$
Solution : Given: $\operatorname{cosec} \theta + \cot \theta = m$ ........(i) Solution: $\operatorname{cosec}^{2}\theta-\cot^{2}\theta = 1$ (Trigonometric identity) $⇒(\operatorname{cosec} \theta - \cot \theta)(\operatorname{cosec} \theta + \cot \theta) =1$ $⇒(\operatorname{cosec} \theta - \cot \theta)=\frac{1}{m}$ ...........(ii) Adding (i) and (ii), we get, $\operatorname{cosec} \theta + \cot \theta + \operatorname{cosec} \theta -\cot \theta = m + \frac{1}{m}$ $⇒2\operatorname{cosec} \theta = \frac{m^{2} + 1}{m}$ $⇒\operatorname{cosec} \theta = \frac{m^{2} + 1}{2m}$ Subtracting (i) and (ii), we get, $\operatorname{cosec} \theta + \cot \theta- \operatorname{cosec} \theta + \cot \theta = m - \frac{1}{m}$ $⇒2\cot \theta = \frac{m^{2} - 1}{m}$ $⇒\cot \theta = \frac{m^{2} - 1}{2m}$ $\therefore \frac{m^{2}–1}{m^{2} + 1}= \frac{m^{2} – 1}{2m}×\frac{2m}{m^2+1}=\cot\theta×\frac{1}{\operatorname{cosec} \theta}=\frac{\frac{\cos \theta}{\sin \theta}}{\frac{1}{\sin \theta}}=\cos \theta$ Hence, the correct answer is $\cos\theta$.
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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 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 : If $\operatorname{cosec} \theta+\cot \theta=p$, then the value of $\frac{p^2-1}{p^2+1}$ 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 : Find $\frac{1}{ \sin \theta}-\sin \theta$.
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