If $x\sin{ heta = y\cos{ heta = \frac{2z an heta}{1 - an^{2} heta}}}$, then $4z^{2}(x^{2} + y^{2})$ is equal to

Question

If $x\sin{	heta = y\cos{	heta = \frac{2z	an	heta}{1 - 	an^{2}	heta}}}$, then $4z^{2}(x^{2} + y^{2})$ is equal to

Dr. Abubacker · Accepted Answer

(c)
Given, $\frac{x}{cosec\ 	heta} = \frac{y}{\sec	heta} = \frac{z}{\cot{2	heta}} = k$ [say]
$	herefore\ \ \ 4z^{2}\left( x^{2} + y^{2} ight) = 4k^{2}\cot^{2}{2	heta(k^{2}cosec^{2}	heta + k^{2}\sec^{2}	heta)}$
$= 4k^{4}\cot^{2}{2	heta\left( \frac{1}{\sin^{2}{	heta\cos^{2}	heta}} ight)}$ $\lbrack\because\sin^{2}{	heta + \cos^{2}{	heta = 1}}brack$
$= \frac{4k^{4}}{4}\left( \frac{\cos^{2}{	heta - \sin^{2}	heta}}{\sin^{2}{	heta\cos^{2}	heta}} ight)^{2}$
$= \left( k^{2}cosec^{2}	heta - k^{2}\sec^{2}	heta ight)^{2}$
$= \left( x^{2} - y^{2} ight)^{2}$

Q. If $x\sin{\theta = y\cos{\theta = \frac{2z\tan\theta}{1 - \tan^{2}\theta}}}$, then $4z^{2}(x^{2} + y^{2})$ is equal to

Explanation

Reviewed by Dr. Abubacker

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