If cos theta = frac{x^{2}-y^{2}}{x^{2}+y^{2}} | vedclass

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(D) Given $\cos 	heta = \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$.Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = \sqrt{1 - \cos

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$\frac{x^{2}-y^{2}}{2xy}$

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(D) Given $\cos 	heta = \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$.Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we have $\sin 	heta = \sqrt{1 - \cos^2 	heta}$.$\sin 	heta = \sqrt{1 - \left(\frac{x^2 - y^2}{x^2 + y^2}ight)^2} = \sqrt{\frac{(x^2 + y^2)^2 - (x^2 - y^2)^2}{(x^2 + y^2)^2}}$.Using $(a+b)^2 - (a-b)^2 = 4ab$,we get $\sin 	heta = \sqrt{\frac{4x^2y^2}{(x^2 + y^2)^2}} = \frac{2xy}{x^2 + y^2}$.Now,$\cot 	heta = \frac{\cos 	heta}{\sin 	heta} = \frac{(x^2 - y^2) / (x^2 + y^2)}{(2xy) / (x^2 + y^2)}$.$\cot 	heta = \frac{x^2 - y^2}{2xy}$.

If $\cos \theta = \frac{x^{2}-y^{2}}{x^{2}+y^{2}}$,then the value of $\cot \theta$ is equal to (where $0^{\circ} \leq \theta \leq 90^{\circ}$)

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