A) 2
B) ?2
C) ½
D) ?1/2
Correct Answer: B
Solution :
\[y={{\tan }^{-1}}(\sec x-\tan x)\] \[\frac{dy}{dx}=\frac{1}{1+{{(\sec x-\tan x)}^{2}}}(\sec x\tan x-{{\sec }^{2}}x)\] \[\frac{dy}{dx}=\frac{{{\cos }^{2}}x.{{\sec }^{2}}x(\sin x-1)}{{{(1-\sin x)}^{2}}+{{\cos }^{2}}x}\] \[\frac{dy}{dx}=\frac{\sin x-1}{1-2\sin x+{{\sin }^{2}}x+{{\cos }^{2}}x}=\frac{\sin x-1}{2(1-\sin x)}=-\frac{1}{2}.\]
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