If sqrt{frac{1+cos A}{1-cos A}}=frac{x}{y},th | vedclass

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(C) Given,$\sqrt{\frac{1+\cos A}{1-\cos A}}=\frac{x}{y}$.Using the identities $1+\cos A = 2\cos^{2}\frac{A}{2}$ and $1-\cos A = 2\sin^{2}\frac{A}{2}$,

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Given,$\sqrt{\frac{1+\cos A}{1-\cos A}}=\frac{x}{y}$.Using the identities $1+\cos A = 2\cos^{2}\frac{A}{2}$ and $1-\cos A = 2\sin^{2}\frac{A}{2}$,we get:$\sqrt{\frac{2\cos^{2}\frac{A}{2}}{2\sin^{2}\frac{A}{2}}} = \frac{x}{y}$$\cot\frac{A}{2} = \frac{x}{y} \implies 	an\frac{A}{2} = \frac{y}{x}$.Now,using the formula $	an A = \frac{2	an\frac{A}{2}}{1-	an^{2}\frac{A}{2}}$:$	an A = \frac{2(\frac{y}{x})}{1-(\frac{y}{x})^{2}} = \frac{\frac{2y}{x}}{\frac{x^{2}-y^{2}}{x^{2}}}$$	an A = \frac{2y}{x} 	imes \frac{x^{2}}{x^{2}-y^{2}} = \frac{2xy}{x^{2}-y^{2}}$.

If $\sqrt{\frac{1+\cos A}{1-\cos A}}=\frac{x}{y}$,then the value of $\tan A$ is equal to

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