NIMCET 2020 Question Paper with solutions

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Question : 48 of 120

Marks: +1, -0

The value of

2\tan^{-1}[\cosec(\tan^{-1}x)-\tan(\cot^{-1}x)]

is

$\tan x$
$\cot x$
$\tan^{-1}x$
$\cosec^{-1}x$

Solution:

\cos a^{-1}(\tan x)\;\;\; \cot^{-1}(\tan x)=1/x

2\tan^{-1}[{√1-x^2}/x-1/x]=2\tan^{-1}[{(1+x^2)-1}/{x[√{1+x^2}+1]}] =2\tan^{-1}[x/{√{1+x^2}+1}]

Put

x=\tan θ

2\tan^{-1}[{\tanθ}/{\sec θ+1}]=2\tan^{-1}[{\sin θ}/{\cos θ}/{1/{\cos θ}+1}]

=2\tan^{-1}[{\tan θ}/{1+\cos θ}]=2\tan^{-1}[{2\sinθ/2\cosθ/2}/{1+2cos^2 θ/2-1}]

=2\tan^{-1}(\tan θ/2)=θ=\tan^{-1}x

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