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JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1
If a1,a2,a3,…an{a_1},{a_2},{a_3}, \ldots {a_n} are in AP with common difference 5 and if aiaj≠−1{a_i}{a_j} \ne - 1 for i,  j=1,  2,  …,ni,\;j = 1,\;2,\; \ldots ,n then

tan⁡−1(51+a1a2)+tan⁡−1(51+a2a3)+...+tan⁡−1(51+an−1an){\tan ^{ - 1}}\left( {\frac{5}{{1 + {a_1}{a_2}}}} \right) + {\tan ^{ - 1}}\left( {\frac{5}{{1 + {a_2}{a_3}}}} \right) + ... + {\tan ^{ - 1}}\left( {\frac{5}{{1 + {a_{n - 1}}{a_n}}}} \right) is equal to
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Correct option: C
tan⁡−1(5n−51+ana1){\tan ^{ - 1}}\left( {\frac{{5n - 5}}{{1 + {a_n}{a_1}}}} \right)
Option C (tan⁡−1(5n−51+ana1){\tan ^{ - 1}}\left( {\frac{{5n - 5}}{{1 + {a_n}{a_1}}}} \right)) is correct because by using the properties of arithmetic progressions and inverse trigonometric functions, the sum simplifies to the given expression.

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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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