JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Analyzing +4 | -1

Let

S

be the set of all solutions of the equation

\cos ^{-1}(2 x)-2 \cos ^{-1}\left(\sqrt{1-x^{2}}\right)=\pi, x \in\left[-\frac{1}{2}, \frac{1}{2}\right] \text {. }

Then

\sum\limits_{{x \in S}} 2 \sin ^{-1}\left(x^{2}-1\right)

is equal to

View answer and explanation

Correct option: A

\frac{-2 \pi}{3}

Option A (

\frac{{-2 \pi}}{{3}}

) is correct because solving the equation and substituting the result into the expression yields this value.

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\sin \left(2 \tan ^{-1}(1 / 3)\right)+

\cos \left(\tan ^{-1} 2 \sqrt{2}\right)

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\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1

x

{x^2} + {y^2} + {z^2} = {r^2},

{\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\;

2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)