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JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Hard

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Analyzing +4 | -1

If

\sin ^{-1} \frac{\alpha}{17}+\cos ^{-1} \frac{4}{5}-\tan ^{-1} \frac{77}{36}=0,0<\alpha<13

, then

\sin ^{-1}(\sin \alpha)+\cos ^{-1}(\cos \alpha)

is equal to

View answer and explanation

Correct option: A

\pi

Option A (

\pi

) is correct because by solving the given equation for

\alpha

, we find

\alpha = 8

, and then

\sin^{-1}(\sin 8) + \cos^{-1}(\cos 8)

simplifies to

\pi

.

More Questions from INVERSE TRIGONOMETRY

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\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)