TS ICET 23 May 2018 Shift 1 Paper

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Question : 135 of 200

Marks: +1, -0

y =

\tan^{-1} [{1+\tan \x}/{1 - \tan \x}]

⇒

{\dy}/{\dx}

=

1
2
0
-1

Solution:

Given

y=\tan^{-1}({1+\tan \x}/{1 - \tan \x})

⇒ \tan y=\;{\tan 45+\tan x}/{1-\tan 45 × \tan x}=\tan (45+x)

⇒y = 45 + x

Differentiating on both the sides, we get

\;{d y}/{d x}=\;{d}/{d x}(45+x)=1

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