Correct Answer: 0.5

Solution : Given expression,
$(\frac{\sec \theta-1}{\sec \theta+1})^n=\operatorname{cosec} \theta-\cot \theta$
Considering L.H.S.,
We know, $\sec\theta=\frac{1}{\cos\theta}$
= $\left\{\frac{\frac{1}{\cos\theta}-1}{\frac{1}{\cos\theta}+1}\right\}^n$
= $\left\{\frac{1-\cos\theta}{1+\cos\theta}\right\}^n$
Multiplying and dividing by $1-\cos\theta$, we get,
= $\left\{\frac{1-\cos\theta}{1+\cos\theta}\times\frac{1-\cos\theta}{1-\cos\theta}\right\}^n$
= $\left\{\frac{(1-\cos\theta)^2}{1-\cos^2\theta}\right\}^n$
= $\left\{\frac{(1-\cos\theta)^2}{\sin^2\theta}\right\}^n$
Now consider R.H.S,
$\operatorname{cosec} \theta-\cot \theta$
We know, $\operatorname{cosec} \theta=\frac{1}{\sin\theta}$ and $\cot\theta=\frac{\cos\theta}{\sin\theta}$
= $\frac{1}{\sin\theta}-\frac{\cos\theta}{\sin\theta}$
= $\frac{1-\cos\theta}{\sin\theta}$
Equating with L.H.S, we get,
⇒ $2n=1$
$\therefore n=0.5$
Hence, the correct answer is 0.5.