JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

{e^{[{{\sin }^2}\alpha + {{\sin }^4}\alpha + {{\sin }^6}\alpha + ...\infty ]{{\log }_e}2}}

is a root of equation

{x^2} - 9x + 8 = 0,

where 0<

\alpha < \frac{\pi }{2},

then the principle value of

{\sin ^{ - 1}}\sin \left( {\frac{{2\pi }}{3}} \right)

View answer and explanation

Correct option: A

\alpha

Option A (

\alpha

) is correct because, after simplifying the given expression and solving for

\alpha

, the principal value of

{\sin ^{ - 1}}\sin \left( {\frac{{2\pi }}{3}} \right)

is equal to

\alpha

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)