Question

If $x \in\left(\frac{\pi}{2}, \pi\right)$, then $\frac{ sec \,x-1}{sec\, x+1}$ if equal to

• A. $\left(\text{cosec}\, x + cot\, x\right)^{2}$
• B. $\left(\sin \, x - cos \, x\right)^{2}$
• C. $\left(\text{cosec} \, x - cot \, x\right)^{2}$
• D. $\left(\sec \, x + tan \, x\right)^{2}$
• E. $\left(\sec \, x - tan \, x\right)^{2}$

Answer 1

Answer: (C)

Given, $ x \in\left(\frac{\pi}{2}, \pi\right)$
$ \therefore \frac{\sec x-1}{\sec x+1}=\frac{\frac{1}{\cos x}-1}{\frac{1}{\cos x}+1} $
$= \frac{1-\cos x}{1+\cos x} \times \frac{1-\cos x}{1-\cos x} $
$= \frac{1^{2}-2 \cos x+\cos ^{2} x}{1-\cos ^{2} x} $
$= \frac{1-2 \cos x+\cos ^{2} x}{\sin ^{2} x} $
$= \operatorname{cosec}^{2} x-2 \cot x \operatorname{cosec} x+\cot ^{2} x $
$=(\operatorname{cosec} x-\cot x)^{2}$