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JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Hard
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Analyzing +4 | -1
Statement I: The equation (sin⁡−1x)3+(cos⁡−1x)3−aπ3=0\left(\sin ^{-1} \mathrm{x}\right)^{3}+\left(\cos ^{-1} \mathrm{x}\right)^{3}-\mathrm{a} \pi^{3}=0 has a solution for all a≥132\mathrm{a} \geq \frac{1}{32}.

Statement II: For any x∈R,sin⁡−1x+cos⁡−1x=π2x \in R, \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} and 0≤(sin⁡−1x−π4)2≤9π2160 \leq\left(\sin ^{-1} \mathrm{x}-\frac{\pi}{4}\right)^{2} \leq \frac{9 \pi^{2}}{16}
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Correct option: A
Both statements I and II are true.
Statement I and II are both true because statement II correctly states the relationship between inverse sine and cosine, and statement I is true because the given equation has a solution for a within the specified range.

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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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