JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Analyzing +4 | -1

The number of real solution of

\cot ^{-1} \sqrt{x(x+4)}+\cos ^{-1} \sqrt{x^2+4 x+1}=\frac{\pi}{2}

is equal to

View answer and explanation

Correct option: C

Option C (2) is correct because the equation is satisfied only when

x^2 + 4x = 0

, leading to two real solutions: 0 and -4.

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

\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)