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JEE Mains · Maths · STD 12 - 2. inverse trigonometric function

यदि \(0 < x < \frac{1}{\sqrt{2}}\) तथा \(\frac{\sin ^{-1} x}{\alpha}=\frac{\cos ^{-1} x}{\beta}\) हैं, तो \(\sin \left(\frac{2 \pi \alpha}{\alpha+\beta}\right)\) का एक मान है

  1. A \(4 \sqrt{\left(1-x^{2}\right)}\left(1-2 x^{2}\right)\)
  2. B \(4 x \sqrt{\left(1-x^{2}\right)}\left(1-2 x^{2}\right)\)
  3. C \(2 x \sqrt{\left(1-x^{2}\right)}\left(1-4 x^{2}\right)\)
  4. D \(4 \sqrt{\left(1-x^{2}\right)}\left(1-4 x^{2}\right)\)
Verified Solution

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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