JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Analyzing +4 | -1

Let

\,{S_K} = \sum\limits_{r = 1}^K {{{\tan }^{ - 1}}\left( {\frac{{{6^r}}}{{{2^{2r + 1}} + {3^{2r + 1}}}}} \right)}

. Then

\,\mathop {\lim \,\,\,}\limits_{K \to \infty } \,{S_K}

is equal to:

View answer and explanation

Correct option: B

{\cot ^{ - 1}}\left( {\frac{3}{2}} \right)

Option B (

{\cot ^{ - 1}}\left( {\frac{3}{2}} \right)

) is correct because the summation telescopes, and the limit is found using inverse trigonometric identities.

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\sin \left(\sin ^{-1} \frac{1}{5}+\cos ^{-1} x\right)=1

x

{x^2} + {y^2} + {z^2} = {r^2},

{\tan ^{ - 1}}\left( {\frac{{xy}}{{zr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{yz}}{{xr}}} \right) + {\tan ^{ - 1}}\left( {\frac{{xz}}{{yr}}} \right)\;

2{\tan ^{ - 1}}\left( {\frac{1}{3}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{7}} \right)