JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

\alpha=\frac{1}{3} \sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\frac{1}{3} \cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)

where

x \geq \frac{4}{3}

, then the value of

\frac{\cos 2 \alpha+\sec \alpha+3 \sqrt{3}}{\sqrt{3}}

is equal to:

View answer and explanation

Correct option: D

\left(\frac{\sqrt{3}}{2}+3\right)

Option D (

\left(\frac{\sqrt{{3}}}{{2}}+3\right)

) is correct because after simplifying the expression and substituting

\alpha = \frac{\pi}{3}

, the value of the expression is

\frac{\sqrt{{3}}}{{2}}+3

More Questions from INVERSE TRIGONOMETRY

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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)