Differentiation Question 287
Question: $ \frac{d}{dx}[ ( \frac{{{\tan }^{2}}2x-{{\tan }^{2}}x}{1-{{\tan }^{2}}2x{{\tan }^{2}}x} )\cot 3x ] $
[AMU 2000]
Options:
A) $ \tan 2x\tan x $
B) $ \tan 3x\tan x $
C) $ {{\sec }^{2}}x $
D) $ \sec x\tan x $
Show Answer
Answer:
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 $ .
BETA
