TS EAMCET · Maths · Inverse Trigonometric Functions
The trigonometric equation \(\sin ^{-1} x=2 \sin ^{-1} a\), has a solution
- A only when \(\frac{1}{\sqrt{2}} \lt a \lt \frac{1}{2}\)
- B for all real values of ' \(a\) '
- C only when \(|a| \leqslant \frac{1}{\sqrt{2}}\)
- D only when \(|a| \geq \frac{1}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(C) only when \(|a| \leqslant \frac{1}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(\sin ^{-1} x=2 \sin ^{-1} a\) will have a solution if \(2 \sin ^{-1} a \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right] \Rightarrow \sin ^{-1} a \in\left[\frac{-\pi}{4}, \frac{\pi}{4}\right]\)…
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