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

Question

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

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

\[ \text{then } k= \]

Using identity

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

\[ =\cos^2x \]

Therefore,

\[ \sqrt{(1+\sin x)(1-\sin x)} = \sqrt{\cos^2x} = |\cos x| \]

Since

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

\(x\) lies in second quadrant, so

\[ \cos x<0 \]

Hence,

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

Now,

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

Comparing with

\[ k\sec x \]

we get

\[ k=-1 \]

\[ \boxed{-1} \]

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