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JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

The value of

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

is

View answer and explanation

Correct option: D

1

Option D (1) is correct because

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

simplifies to

\sin(\frac{\pi}{3} + \frac{\pi}{6})

, which equals

\sin(\frac{\pi}{2})

or 1.

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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},

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2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)