JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Application +4 | -1

\sum_{\mathrm{r}=1}^{20} \tan ^{-1}\left(\frac{1}{2 \mathrm{r}^2}\right)=\tan ^{-1} \frac{\mathrm{m}}{\mathrm{n}}

; where

\mathrm{m}

and

\mathrm{n}

are co-prime numbers then

\left[\log _2(m+n)\right]

equals (where [.] is greatest integer function)

View answer and explanation

Correct option: A

Option A (15) is correct because

\cot 7 \frac{1^{\circ}}{2} = \sqrt{6} + \sqrt{4} + \sqrt{3} + \sqrt{2}

, and the sum of the numbers under the square roots is 15.

More Questions from INVERSE TRIGONOMETRY

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\cos ^{-1}(\cos 12)-\sin ^{-1}(\sin 12)

\sin \left(2 \tan ^{-1}(1 / 3)\right)+

\cos \left(\tan ^{-1} 2 \sqrt{2}\right)

\cos ^ {-1}\left\{\frac{1}{\sqrt{2}}\left(\cos \frac{9 \pi}{10}-\sin \frac{9 \pi}{10}\right)\right\}=

\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)