Question

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Answer 1

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Answer 2

Given :

$\sf \bullet \: x \: = \: a\cos(\theta) \: + \: b\sin(\theta)$

$\sf \bullet \: y \: = \: a\sin(\theta) \: - \: b\cos(\theta)$

To prove :

x² + y² = a² + b²

Solution :

We have ,

• LHS = x² + y²
• RHS = a² + b²

Now what we have to do is prove LHS = RHS

Therefore firstly solving LHS

$\sf \implies \: LHS \: = \: x^2 \: + \: {y}^{2}$

Putting value of x and y

$\sf \implies \: LHS \: = \: \bigg( a\cos(\theta) \: + \: b\sin(\theta) \bigg)^2 + \bigg( a\sin(\theta) \: - \: b\cos(\theta) \bigg)^2$

$\sf \implies \: LHS \: = \: \bigg( a^2\cos^2(\theta) \: + \: b^2\sin^2(\theta) + 2[a\cos(\theta)][b\sin(\theta)] \bigg) + \bigg( a^2\sin^2(\theta) \: + \: b^2\cos^2(\theta) - 2[a\sin(\theta)][b\cos(\theta)] \bigg)$

$\sf \implies \: LHS \: = \: a^2\cos^2(\theta) \: + a^2\sin^2(\theta) \: + \: b^2\sin^2(\theta) \: + \: b^2\cos^2(\theta) \:+\: 2[a\cos(\theta)][b\sin(\theta)] </p><p> \: - 2[a\sin(\theta)][b\cos(\theta)]$

$\sf \implies \: LHS \: = \: a^2[\cos^2(\theta) \: + \sin^2(\theta)] \: + \: b^2[\sin^2(\theta) \: + \: \cos^2(\theta)] \:+\: 2ab\cos(\theta)\sin(\theta) </p><p> \: - 2ab\sin(\theta)\cos(\theta)]$

$\sf \implies \: LHS \: = \: a^2 \: + \: b^2 \:$

$\sf \implies \: LHS \: = RHS$

Hence proved