SSC CPO 2024 PYQ — Question 135

The correct answer is option 3 i.e. cot²θ.

$\frac{1 - \cot^2 \theta}{\tan^2 \theta - 1}$

Use the identity for cotangent and tangent
$\cot \theta = \frac{1}{\tan \theta}$

Therefore, $\cot^2 \theta = \frac{1}{\tan^2 \theta}$

Substitute $\cot^2 \theta = \frac{1}{\tan^2 \theta}$ into the given expression:
$\frac{1 - \frac{1}{\tan^2 \theta}}{\tan^2 \theta - 1}$

The numerator is $1 - \frac{1}{\tan^2 \theta}$, which can be written as:
⇒ $\frac{\tan^2 \theta - 1}{\tan^2 \theta}$

⇒ $\frac{\frac{\tan^2 \theta - 1}{\tan^2 \theta}}{\tan^2 \theta - 1}$

Since $\tan^2 \theta - 1$ appears in both the numerator and the denominator, we can cancel them out (as long as $\tan^2 \theta \neq 1$):
⇒ $\frac{1}{\tan^2 \theta}$

We know that $\frac{1}{\tan^2 \theta} = \cot^2 \theta$.
⇒ $\frac{1 - \cot^2 \theta}{\tan^2 \theta - 1} = \cot^2 \theta$