JEE Main
Exam JEE Main
Subject Mathematics
Chapter Inverse Trigonometric Functions JEE Main · Mathematics · Inverse Trigonometric Functions
If y= ^ -1 ( x+ x), then (d y)/(d x)=
If
y = tan − 1 ( sec x + tan x ) , y=\tan ^{-1}(\sec x+\tan x), \quad y = tan − 1 ( sec x + tan x ) , then
d y d x = \frac{d y}{d x}= d x d y = Options A 1 2 \frac{1}{2} 2 1 B 1 C − 1 2 \frac{-1}{2} 2 − 1 D − 1 -1 − 1 Correct answer A. 1 2 \frac{1}{2} 2 1
Step-by-step solution
Given
y = tan − 1 ( 1 cos x + sin x cos x ) = tan − 1 ( 1 + sin x cos x ) = tan − 1 [ ( cos 2 x 2 + sin 2 x 2 ) + 2 sin x 2 cos x 2 cos 2 x 2 − sin 2 x 2 ) = tan − 1 { x ( cos x 2 + sin x 2 ) ( cos x 2 − sin x 2 ) } = tan − 1 { cos x 2 + sin x 2 cos x 2 − sin x 2 } = tan − 1 ( 1 + tan x 2 1 − tan x 2 ) = tan − 1 ( tan ( π 4 + x 2 ) ) \begin{aligned} y &=\tan ^{-1}\left(\frac{1}{\cos x}+\frac{\sin x}{\cos x}\right)=\tan ^{-1}\left(\frac{1+\sin x}{\cos x}\right) \\ &=\tan ^{-1}\left[\frac{\left(\cos ^{2} \frac{x}{2}+\sin ^{2} \frac{x}{2}\right)+2 \sin \frac{x}{2} \cos \frac{x}{2}}{\cos ^{2} \frac{x}{2}-\sin ^{2} \frac{x}{2}}\right) \\ &=\tan ^{-1}\left\{\frac{x}{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}\right\} \\ &=\tan ^{-1}\left\{\frac{\cos \frac{x}{2}+\sin \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}\right\}=\tan ^{-1}\left(\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}\right) \\ &=\tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right) \end{aligned} y = tan − 1 ( cos x 1 + cos x sin x ) = tan − 1 ( cos x 1 + sin x ) = tan − 1 [ cos 2 2 x − sin 2 2 x ( cos 2 2 x + sin 2 2 x ) + 2 sin 2 x cos 2 x ) = tan − 1 { ( cos 2 x + sin 2 x ) ( cos 2 x − sin 2 x ) x } = tan − 1 { cos 2 x − sin 2 x cos 2 x + sin 2 x } = tan − 1 ( 1 − tan 2 x 1 + tan 2 x ) = tan − 1 ( tan ( 4 π + 2 x ) ) y = π 4 + x 2 ∴ d y d x = 0 + 1 2 = 1 2 \begin{aligned} \mathrm{y} &=\frac{\pi}{4}+\frac{\mathrm{x}}{2} \\ \therefore \frac{\mathrm{dy}}{\mathrm{dx}} &=0+\frac{1}{2}=\frac{1}{2} \end{aligned} y ∴ dx dy = 4 π + 2 x = 0 + 2 1 = 2 1 $
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