JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

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Analyzing +4 | -1

Let

s=\left\{x \in \mathrm{R}: 0 < x < 1\right.

and

\left.2 \tan ^{-1}\left(\frac{1-x}{1+x}\right)=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\right\}

n(S)

denotes the number of elements in

S

then:

View answer and explanation

Correct option: C

n(S)=1

and the element in

S

is less than

\frac{1}{2}

Option C is correct because solving the given equation using trigonometric identities yields a single solution for x, which is less than 1/2.

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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)