Question

\[ \text{If } \pi<x<2\pi, \]

\[ \sqrt{(1+\cos x)(1-\cos x)} + \sqrt{(1-\cos x)(1+\cos x)} = k\cosec x \]

\[ \text{then } k= \]

Solution

Using identity

\[ (1+\cos x)(1-\cos x)=1-\cos^2x \]

\[ =\sin^2x \]

Therefore,

\[ \sqrt{\sin^2x}+\sqrt{\sin^2x} = |\sin x|+|\sin x| \]

\[ =2|\sin x| \]

Since

\[ \pi<x<2\pi \]

\(x\) lies in III or IV quadrant where

\[ \sin x<0 \]

Hence,

\[ |\sin x|=-\sin x \]

\[ 2|\sin x| = -2\sin x \]

Now,

\[ -2\sin x = -2\cdot\frac1{\cosec x} \]

Comparing with

\[ k\cosec x \]

we get

\[ k=-2 \]

Answer

\[ \boxed{-2} \]