Differentiation Question 378
Question: $ \frac{d}{dx}\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}= $
[AISSE 1985; DSSE 1986]
Options:
A) $ {{\sec }^{2}}x $
B) $ -{{\sec }^{2}}( \frac{\pi }{4}-x ) $
C) $ {{\sec }^{2}}( \frac{\pi }{4}+x ) $
D) $ {{\sec }^{2}}( \frac{\pi }{4}-x ) $
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Answer:
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 ( \frac{\pi }{4}-x )\Rightarrow \frac{dy}{dx}=-{{\sec }^{2}}( \frac{\pi }{4}-x ) $ .
BETA
