A) \[{{\sec }^{2}}x\]
B) \[-{{\sec }^{2}}\left( \frac{\pi }{4}-x \right)\]
C) \[{{\sec }^{2}}\left( \frac{\pi }{4}+x \right)\]
D) \[{{\sec }^{2}}\left( \frac{\pi }{4}-x \right)\]
Correct Answer: B
Solution :
Let \[y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}\] \[=\sqrt{\frac{{{(\cos x-\sin x)}^{2}}}{{{(\cos x+\sin x)}^{2}}}}\] \[=\frac{1-\tan }{1+\tan x}\] \[\Rightarrow \] \[y=\tan \left( \frac{\pi }{4}-x \right)\] On differentiating both sides w.r.t. \[x,\]we get \[\frac{dy}{dx}=-{{\sec }^{2}}\left( \frac{\pi }{4}-x \right)\]
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