A) \[\tan 2x\,\tan x\]
B) \[\tan 3x\tan x\]
C) \[{{\sec }^{2}}x\]
D) \[\sec x\tan x\]
Correct Answer: C
Solution :
Let \[y=\frac{{{\tan }^{2}}2x-{{\tan }^{2}}x}{1-{{\tan }^{2}}2x{{\tan }^{2}}x}\] = \[\frac{(\tan 2x-\tan x)}{(1+\tan 2x\tan x)}\,\frac{(\tan 2x+\tan x)}{(1-\tan 2x\tan x)}\] = \[\tan (2x-x)\,\tan (2x+x)\] = \[\tan x\tan 3x\]. \ \[\frac{d}{dx}[y.\cot 3x]=\frac{d}{dx}[\tan x]={{\sec }^{2}}x\].You need to login to perform this action.
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