JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
If \(0 < x < \frac{1}{\sqrt{2}}\) and \(\frac{\sin ^{-1} x}{\alpha}=\frac{\cos ^{-1} x}{\beta}\), then a value of \(\sin \left(\frac{2 \pi \alpha}{\alpha+\beta}\right)\) is\(......\)
- A \(4 \sqrt{\left(1-x^{2}\right)}\left(1-2 x^{2}\right)\)
- B \(4 x \sqrt{\left(1-x^{2}\right)}\left(1-2 x^{2}\right)\)
- C \(2 x \sqrt{\left(1-x^{2}\right)}\left(1-4 x^{2}\right)\)
- D \(4 \sqrt{\left(1-x^{2}\right)}\left(1-4 x^{2}\right)\)
Answer & Solution
Correct Answer
(B) \(4 x \sqrt{\left(1-x^{2}\right)}\left(1-2 x^{2}\right)\)
Step-by-step Solution
Detailed explanation
\(\frac{\sin ^{-1} x}{\alpha}=\frac{\cos ^{-1} x}{\beta}= k\) \(\sin ^{-1} x=k \alpha\) \(\cos ^{-1} x=k \beta\) \(k=\frac{\pi}{2(\alpha+\beta)}\) \(\sin \left(\frac{2 \pi \alpha}{\alpha+\beta}\right)=\sin \left(4 \sin ^{-1} x\right)\)…
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