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JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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If A   and   BA\,\;\mathrm{and}\;\,B are acute positive angles satisfying the equations 3  sin⁡2A+2  sin⁡2  B=1   and   3  sin⁡  2A−2  sin⁡  2B=03\,\,{\sin ^2}A + 2\,\,{\sin ^2}\,\,B = 1\,\;\mathrm{and}\;\,3\,\,\sin \,\,2A - 2\,\,\sin \,\,2B = 0, then A+2BA + 2B is equal to
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Correct option: B
π2\frac{\pi }{2}
Option B (π2\frac{\pi}{2}) is correct because, by manipulating the given trigonometric equations and using trigonometric identities, we can find that A+2B=π2A + 2B = \frac{\pi}{2}.

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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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