JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Application +4 | -1

f\left( x \right) = {\sin ^{ - 1}}\left\{ {\frac{{\sqrt 3 }}{2}x - \frac{1}{2}\sqrt {1 - {x^2}} } \right\},\; - \frac{1}{2} \le x \le 1

, then

f\left( x \right)

is equal to

View answer and explanation

Correct option: B

{\sin ^{ - 1}}x - \frac{\pi }{6}

Option B (

{\sin ^{ - 1}}x - \frac{\pi }{6}

) is correct because by substituting

x = \sin \theta

, the expression simplifies to

{\sin ^{ - 1}}(\sin(\theta - \frac{\pi}{6}))

, which equals

\theta - \frac{\pi}{6}

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\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)