Question

Question in attachment plz solve it . its urgent

Answer 1

Step-by-step explanation:

We have,

$\cos ^{ - 1} x - \cos^{ - 1} \bigg( \frac{y}{2} \bigg) = \alpha$

$\implies \cos^{ - 1} \bigg( \frac{xy}{2} + \sqrt{1 - {x}^{2} }. \sqrt{1 - \frac{ {y}^{2} }{4} } \bigg) = \alpha$

$\implies \frac{xy}{2} + \sqrt{1 - {x}^{2} }. \sqrt{1 - \frac{ {y}^{2} }{4} } = \cos(\alpha)$

$\implies \sqrt{1 - {x}^{2} }. \sqrt{1 - \frac{ {y}^{2} }{4} } = \cos(\alpha) - \frac{xy}{2}$

$\implies \bigg( \sqrt{1 - {x}^{2} }. \sqrt{1 - \frac{ {y}^{2} }{4} } \bigg) ^{2} = \bigg( \cos(\alpha) - \frac{xy}{2} \bigg) ^{2}$

$\implies (1 - {x}^{2}) \bigg( 1 - \frac{ {y}^{2} }{4} \bigg) = \cos ^{2} (\alpha) + \frac{x^{2} y^{2} }{4} - xy \cos( \alpha )$

$\implies 1 - {x}^{2} - \frac{ {y}^{2} }{4} + \frac{ {x}^{2} {y}^{2} }{4} = \cos ^{2} (\alpha) + \frac{x^{2} y^{2} }{4} - xy \cos( \alpha )$

$\implies 1 - {x}^{2} - \frac{ {y}^{2} }{4} = \cos ^{2} (\alpha) - xy \cos( \alpha )$

$\implies 1 - \cos ^{2} (\alpha) = {x}^{2} +\frac{ {y}^{2} }{4} - xy \cos( \alpha )$

$\implies \sin^{2} (\alpha) = {x}^{2} +\frac{ {y}^{2} }{4} - xy \cos( \alpha )$

$\implies 4 \sin^{2} (\alpha) = 4{x}^{2} +{y}^{2} - 4xy \cos( \alpha )$