JEE Mains · Maths · STD 12 - 2. inverse trigonometric function
If \({\cos ^{ - 1}}\,x\, - \,{\cos ^{ - 1}}\,\frac{y}{2}\, = \,\alpha ,\) where \( - {\kern 1pt} 1\, \le \,x\, \le \,1,\,\) \(- {\kern 1pt} 2\, \le \,y\, \le \,2,\) \(x\, \le \,\,\frac{y}{2},\) then for all \(x, y, 4x^2 -4xy\,\,cos\,\alpha + y^2\) is equal to
- A \(4\,{\sin ^2}\,\alpha \, - \,2{x^2}{y^2}\)
- B \(4\,{\cos ^2}\,\alpha \, + \,2{x^2}{y^2}\)
- C \(2{\sin ^2}\,\alpha \,\)
- D \(4{\sin ^2}\,\alpha \,\)
Answer & Solution
Correct Answer
(D) \(4{\sin ^2}\,\alpha \,\)
Step-by-step Solution
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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