The value $\tan \left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)$ is

JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Hard

1 of 1

Application +4 | -1

The value

\tan \left(\frac{1}{2} \cos ^{-1} \frac{\sqrt{5}}{3}\right)

View answer and explanation

Correct option: B

\frac{3-\sqrt{5}}{2}

Option B ((3-√5)/2) is correct because using the half-angle formula for tangent and simplifying leads to the correct result.

Practice All Chapter MCQs →

\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)