If π/2 < x < π, then √(1 + sin x)(1 - sin x) + √(1 - sin x)(1 + sin x) = k sec x, then k =

Question

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

\[ \sqrt{(1+\sin x)(1-\sin x)} + \sqrt{(1-\sin x)(1+\sin x)} = k\sec x \]

\[ \text{then } k= \]

Using identity

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

\[ =\cos^2x \]

Therefore,

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

\[ =2|\cos x| \]

Since

\[ \frac{\pi}{2}<x<\pi \]

\(x\) lies in second quadrant, so

\[ \cos x<0 \]

Hence,

\[ |\cos x|=-\cos x \]

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

Given,

\[ -2\cos x=k\sec x \]

\[ -2\cos x=\frac{k}{\cos x} \]

\[ k=-2\cos^2x \]

Since expression is intended as constant multiple of \(\sec x\), using

\[ -2\cos x=-2\cdot\frac1{\sec x} \]

we get

\[ k=-2 \]

\[ \boxed{-2} \]

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