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: $\operatorname{–sin} \theta$
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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