NIMCET 2017 Question Paper with solutions

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Question : 84 of 120

Marks: +1, -0

If

sin^{-1}

{2a}/{1+a^2}

+

sin^{-1} {2b}/{1+b^2}

=

2tan^{-1} n

, then

n = $(a-b)/(1+ab)$
n = $(ab)/(a-b)$
n = $(a+b)/(1-ab)$
n = $(1-ab)/(1+ab)$

Solution:

⇒

2tan^{-1}a

+

2tan^{-1}b

=

2ta^{-1}n

⇒

tan^{-1}a+tan^{-1}b

=

tan^{-1}n

⇒

tan^{-1}({a+b}/{1-ab})

=

tan^{-1}

n
⇒

{a+b}/{1-ab}

= n

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