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JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Application +4 | -1

The value of

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

is equal to

View answer and explanation

Correct option: B

105^{\circ}

Option B (

105^{\circ}

) is correct because

\sin^{-1}(\frac{\sqrt{3}}{2}) + \sin^{-1}(\frac{1}{\sqrt{2}})

equals

\frac{\pi}{3} + \frac{\pi}{4}

, which simplifies to

\frac{7\pi}{12}

or

105^{\circ}

.

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

is- The value of

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

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\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)