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JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Assertion: To define the inverse of the function f(x)=tan⁡x{f}({x})=\tan {x} any of the intervals (−3π2,−π2),(−π2,π2),(π2,3π2)\left(-\frac{3 \pi}{2}, \frac{-\pi}{2}\right),\left(-\frac{\pi}{2}, \frac{\pi}{2}\right),\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right) etc. can be chosen.

Reason: The branch having range (−π2,π2)\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) is called principal value branch of the function g(x)=tan⁡−1xg(x)=\tan ^{-1} x.
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Correct option: B
Assertion is correct, reason is correct; reason is not a correct explanation for assertion
Option B is correct because the assertion is correct, the reason is correct, but the reason is not a correct explanation for the assertion. The assertion is true because tangent is bijective on each interval, and the reason correctly states the principal value branch.

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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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