JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

The value of

\frac{\alpha^2}{2} \operatorname{cosec}^2\left(\frac{1}{2} \tan ^{-1} \frac{\alpha}{\beta}\right)-\frac{\beta^2}{2} \sec ^2\left(\frac{1}{2} \tan ^{-1} \frac{\beta}{\alpha}\right)

is equal to

View answer and explanation

Correct option: D

\sqrt{\alpha^2+\beta^2}(\beta+\alpha)

Option D (

\sqrt{{\alpha^2+\beta^2}}(\beta+\alpha)

) is correct because after simplifying the trigonometric expressions and substituting, the value equals to the given expression.

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\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)