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

यदि \(\cos ^{-1} x -\cos ^{-1} \frac{ y }{2}=\alpha\), जहाँ \(-1 \leq x \leq 1,-2 \leq y \leq 2, x \leq \frac{ y }{2}\) है, तो सभी \(x , y\) के लिए, \(4 x ^{2}-4 xy \cos \alpha+ y ^{2}\) बराबर है

  1. A \(4\,{\sin ^2}\,\alpha \, - \,2{x^2}{y^2}\)
  2. B \(4\,{\cos ^2}\,\alpha \, + \,2{x^2}{y^2}\)
  3. C \(2{\sin ^2}\,\alpha \,\)
  4. D \(4{\sin ^2}\,\alpha \,\)
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Answer & Solution

Correct Answer

(D) \(4{\sin ^2}\,\alpha \,\)

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Detailed explanation

\({\cos ^{ - 1}}x - {\cos ^{ - 1}}\frac{y}{2} = \alpha \) \(\cos \left( {{{\cos }^{ - 1}}x - {{\cos }^{ - 1}}\frac{y}{2}} \right) = \cos \alpha \) \( \Rightarrow x \times \frac{y}{2} + \sqrt {1 - {x^2}} \sqrt {1 - \frac{{{y^2}}}{4}} = \cos \alpha \)…
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