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 :
\[y=\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}=\frac{\cos x-\sin x}{\cos x+\sin x}\] \[=\frac{1-\tan x}{1+\tan x}=\tan \left( \frac{\pi }{4}-x \right)\Rightarrow \frac{dy}{dx}=-{{\sec }^{2}}\left( \frac{\pi }{4}-x \right)\].
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