AP EAMCET · Maths · Inverse Trigonometric Functions
If \(\sin ^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\ldots \infty\right)\)
\(+\cos ^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \infty\right)=\frac{\pi}{2}\) and \(0 < \mathrm{x} < \sqrt{2}\),
then \(x\) is equal to
- A \(\frac{1}{2}\)
- B \(1\)
- C \(\frac{-1}{2}\)
- D \(-1\)
Answer & Solution
Correct Answer
(B) \(1\)
Step-by-step Solution
Detailed explanation
Given that, \(\sin ^{-1}\left(x-\frac{x^2}{2}+\frac{x^2}{4}-\ldots \infty\right)\) \(\cos ^{-1}\left(x-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \infty\right)=\frac{\pi}{2}\) Here, \(\quad x-\frac{x^2}{2}+\frac{x^3}{4}-\ldots \infty\) Forms G.P. so its sum can be given as…
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