JEE MATHEMATICS MCQ: INVERSE TRIGONOMETRY

1 Easy

1 of 1

Application +4 | -1

\sin ^{-1}\left(x-\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

equal to

View answer and explanation

Correct option: A

Option A (1) is correct. The equation simplifies to x = x², leading to x = 1 as the only valid solution within the given range.

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x

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

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