If $\sin ^{-1}\left(\frac{6 x}{1+9 x^2}\right)=2 \tan ^{-1}(a x)$, then $a=$

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

\sin ^{-1}\left(\frac{6 x}{1+9 x^2}\right)=2 \tan ^{-1}(a x)

, then

a=

View answer and explanation

Correct option: A

Option A (3) is correct because by using the double angle formula for sine and comparing the expressions, a is found to be 3.

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