JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Hard

1 of 1

Application +4 | -1

\sin ^{-1}\left(-\frac{x^2}{2}+\frac{x^3}{4}-\ldots\right)+\cos ^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots\right)

=\frac{\pi}{2}

for

0<|x|<\sqrt{2}

, then

x

equals

View answer and explanation

Correct option: B

Option B (1) is correct because simplifying the given equation using the properties of inverse trigonometric functions and geometric series leads to x = 1.

More Questions from INVERSE TRIGONOMETRY

Practice All Chapter MCQs →

The value of

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

is- The value of

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

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

is Find

\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

, then

x

is equal to If

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

then

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

is equal to

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

is equal to