Question

If cos-1(x/a) + cos-1(y/b) = alpha,
Prove that, x2/a2 + y2/b2 - 2xy/ab * [cos(alpha)]= sin2(alpha)​​

Answer 1

➡HERE IS YOUR ANSWER⬇

${cos}^{ - 1} \frac{x}{a} + {cos}^{ - 1} \frac{y}{b} = \alpha \\ \\ or \: \: {cos}^{ - 1} ( \frac{x}{a} \frac{y}{b} + \sqrt{(1 - \frac{ {x}^{2} }{ {a}^{2} })(1 - \frac{ {y}^{2} }{ {b}^{2} } ) }) = \alpha \\ \\ or \: \: \frac{x}{a} \frac{y}{b} + \sqrt{(1 - \frac{ {x}^{2} }{ {a}^{2} })(1 - \frac{ {y}^{2} }{ {b}^{2} } ) } = cos \alpha \\ \\ or \: \: \sqrt{(1 - \frac{ {x}^{2} }{ {a}^{2} })(1 - \frac{ {y}^{2} }{ {b}^{2} } ) } = cos \alpha - \frac{x}{a} \frac{y}{b} \\ \\ squaring \: \: both \: \: sides \: \: we \: \: get \\ \\ (1 - \frac{ {x}^{2} }{ {b}^{2} } )(1 - \frac{ {y}^{2} }{ {b}^{2} } ) = {(cos \alpha - \frac{xy}{ab} )}^{2} \\ \\ or \: \: 1 - \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } + \frac{ {x}^{2} {y}^{2} }{ {a}^{2} {b}^{2} } = {cos}^{2} \alpha - 2 \frac{xy}{ab} cos \alpha + \frac{ {x}^{2} {y}^{2} }{ {a}^{2} {b}^{2} } \\ \\ cancelling \: \: \frac{ {x}^{2} {y}^{2} }{ {a}^{2} {b}^{2} } \: \: we \: \: get \\ \\ 1 - \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = {cos}^{2} \alpha - 2 \frac{xy}{ab} cos \alpha \\ \\ or \: \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } - 2 \frac{xy}{ab} cos \alpha = 1 - {cos}^{2} \alpha \\ \\ or \: \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } - 2 \frac{xy}{ab} cos \alpha = {sin}^{2} \alpha \: \: (proved)$

⬆HOPE THIS HELPS YOU⬅

Answer 2

Step-by-step explanation:

➡

$\begin{lgathered} \sf \: {cos}^{ - 1} \frac{x}{a} + {cos}^{ - 1} \frac{y}{b} = \alpha \\ \\ \sf \: or \: \: {cos}^{ - 1} ( \frac{x}{a} \frac{y}{b} + \sqrt{(1 - \frac{ {x}^{2} }{ {a}^{2} })(1 - \frac{ {y}^{2} }{ {b}^{2} } ) }) = \alpha \\ \\ \sf \: or \: \: \frac{x}{a} \frac{y}{b} + \sqrt{(1 - \frac{ {x}^{2} }{ {a}^{2} })(1 - \frac{ {y}^{2} }{ {b}^{2} } ) } = cos \alpha \\ \\ \sf \: or \: \: \sqrt{(1 - \frac{ {x}^{2} }{ {a}^{2} })(1 - \frac{ {y}^{2} }{ {b}^{2} } ) } = cos \alpha - \frac{x}{a} \frac{y}{b} \\ \\ \sf \large \green{squaring \: \: both \: \: sides \: \: we \: \: get} \\ \\ \sf \: (1 - \frac{ {x}^{2} }{ {b}^{2} } )(1 - \frac{ {y}^{2} }{ {b}^{2} } ) = {(cos \alpha - \frac{xy}{ab} )}^{2} \\ \\ \sf \: or \: \: 1 - \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } + \frac{ {x}^{2} {y}^{2} }{ {a}^{2} {b}^{2} } = {cos}^{2} \alpha - 2 \frac{xy}{ab} cos \alpha + \frac{ {x}^{2} {y}^{2} }{ {a}^{2} {b}^{2} } \\ \\ \sf \: cancelling \: \: \frac{ {x}^{2} {y}^{2} }{ {a}^{2} {b}^{2} } \: \: we \: \: get \\ \\ \sf \: 1 - \frac{ {x}^{2} }{ {a}^{2} } - \frac{ {y}^{2} }{ {b}^{2} } = {cos}^{2} \alpha - 2 \frac{xy}{ab} cos \alpha \\ \\ \sf \: or \: \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } - 2 \frac{xy}{ab} cos \alpha = 1 - {cos}^{2} \alpha \\ \\ \sf \orange{or \: \: \frac{ {x}^{2} }{ {a}^{2} } + \frac{ {y}^{2} }{ {b}^{2} } - 2 \frac{xy}{ab} cos \alpha = {sin}^{2} \alpha} \: \: (proved)\end{lgathered} \\ </p><p>$