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JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

If

- 1 \le x \le - \frac{1}{2}

, then

{\sin ^{ - 1}}\left( {3x - 4{x^3}} \right)

equals

View answer and explanation

Correct option: C

- \pi - 3{\sin ^{ - 1}}x

Option D (

\sqrt 3

) is correct because the expression simplifies to

\tan(\frac{2\pi}{5} - \frac{\pi}{15}) = \tan(\frac{\pi}{3}) = \sqrt{3}

.

More Questions from INVERSE TRIGONOMETRY

Practice All Chapter MCQs →

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

is- The value of

\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)