JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

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Analyzing +4 | -1

{\cos ^{ - 1}}x + {\cos ^{ - 1}}y = \frac{\pi }{2}\;

and

{\tan ^{ - 1}}x - {\tan ^{ - 1}}y = 0,\;

then

{x^2} + xy + {y^2}\;

is equal to

View answer and explanation

Correct option: C

\frac{3}{2}

Option C (3/2) is correct because from the given equations, we can deduce that x = y = 1/sqrt(2), and substituting these values into the expression x² + xy + y² yields 3/2.

More Questions from INVERSE TRIGONOMETRY

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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