JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

0 < x < \frac{1}{{\sqrt 2 }}

and

\frac{{{{\sin }^{ - 1}}x}}{\alpha } = \frac{{{{\cos }^{ - 1}}x}}{\beta }

then a value of

\sin \left( {\frac{{2\pi \alpha }}{{\alpha + \beta }}} \right)

View answer and explanation

Correct option: B

4 x \sqrt{\left(1-x^2\right)}\left(1-2 x^2\right)

Option B is correct because by using trigonometric identities and the given condition, we can simplify the expression and find the value of

\sin \left( {\frac{{2\pi \alpha }}{{\alpha + \beta }}} \right)

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\cos ^{-1}(\cos 12)-\sin ^{-1}(\sin 12)

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

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

\cos ^ {-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}=

\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)