JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Application +4 | -1

\alpha=\cos ^{-1}\left(\frac{3}{5}\right), \beta=\tan ^{-1}\left(\frac{1}{3}\right)

, where

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

, then

\alpha-\beta

is equal to :

View answer and explanation

Correct option: D

\sin ^{-1}\left(\frac{9}{5 \sqrt{10}}\right)

The correct answer is D (

\sin^{-1}(\frac{9}{5 \sqrt{10}})

) because it is obtained by using the tangent subtraction formula and then converting to arcsin.

More Questions from INVERSE TRIGONOMETRY

Practice All Chapter MCQs →

\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)