JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Application +4 | -1

\tan ^{-1}\left(\frac{2 x}{1-x^{2}}\right)+\cot ^{-1}\left(\frac{1-x^{2}}{2 x}\right)=\frac{\pi}{3}, x \in[-1,1]

sum of all solutions is

\alpha-\frac{4}{\sqrt{3}}

, then

\alpha

is:

View answer and explanation

Correct option: B

Option B (2) is correct because after solving the equation, the sum of all solutions is indeed 2 - 4/sqrt(3), making alpha equal to 2.

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

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\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)