Question

It x= a coso-bsino
y=a sin +bcoso
then prove that a +
b² = x² + y²​

Answer 1

$\textbf{Given:}$

$x=a\:cos\,\theta-b\:sin\,\theta$

$y=a\:sin\,\theta+b\:cos\,\theta$

$\textbf{To find:}$

$a^2+b^2=x^2+y^2$

$\textbf{Solution:}$

$\text{Consider,}$

$x=a\:cos\,\theta-b\:sin\,\theta$

$y=a\:sin\,\theta+b\:cos\,\theta$

$\text{squaring and adding, these two equations, we get}$

$x^2+y^2=(a\:cos\,\theta-b\:sin\,\theta)^2+(a\:sin\,\theta+b\:cos\,\theta)^2$

$x^2+y^2=a^2cos^2\theta+b^2\:sin^2\theta-2ab\,cos\,\theta\;sin\,\theta+a^2sin^2\theta+b^2cos^2\theta-2ab\,cos\,\theta\;sin\,\theta$

$x^2+y^2=a^2cos^2\theta+b^2\:sin^2\theta+a^2sin^2\theta+b^2cos^2\theta$

$x^2+y^2=a^2(cos^2\theta+sin^2\theta)+b^2(sin^2\theta+cos^2\theta)$

$\text{Using}$

$\boxed{\bf,cos^2A+sin^2A=1}$

$x^2+y^2=a^2(1)+b^2(1)$

$\implies\boxed{\bf\,x^2+y^2=a^2+b^2}$

$\text{Hence proved}$

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