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JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1
If sin⁡−1α+sin⁡−1β+sin⁡−1γ=3π2,  {\sin ^{ - 1}}\mathrm{\alpha } + {\sin ^{ - 1}}\mathrm{\beta } + {\sin ^{ - 1}}\mathrm{\gamma } = \frac{{3\pi }}{2},\; then αβ+αγ+βγ\mathrm{\alpha \beta } + \mathrm{\alpha \gamma } + \mathrm{\beta \gamma } is equal to
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Correct option: C
3
Option C (3) is correct because, given sin⁡−1α+sin⁡−1β+sin⁡−1γ=3π2{\sin ^{ - 1}}\mathrm{\alpha } + {\sin ^{ - 1}}\mathrm{\beta } + {\sin ^{ - 1}}\mathrm{\gamma } = \frac{{3\pi }}{2}, it implies α=β=γ=1\alpha = \beta = \gamma = 1, and thus αβ+αγ+βγ=3\mathrm{\alpha \beta } + \mathrm{\alpha \gamma } + \mathrm{\beta \gamma } = 3.

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The value of cos⁡−1(cos⁡12)−sin⁡−1(sin⁡12)\cos ^{-1}(\cos 12)-\sin ^{-1}(\sin 12) is- The value of sin⁡(2tan⁡−1(1/3))+\sin \left(2 \tan ^{-1}(1 / 3)\right)+ cos⁡(tan⁡−122)\cos \left(\tan ^{-1} 2 \sqrt{2}\right) is Find cos⁡−1{12(cos⁡9π10−sin⁡9π10)}=\cos ^ {-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}= If sin⁡(sin⁡−115+cos⁡−1x)=1\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1, then xx is equal to If x2+y2+z2=r2,{x^2} + {y^2} + {z^2} = {r^2}, then

tan⁡−1(xyzr)+tan⁡−1(yzxr)+tan⁡−1(xzyr)  {\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\; is equal to 2tan⁡−1(13)+tan⁡−1(17)2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right) is equal to
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