Prove the following identities, where the ang | vedclass

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Question

$\frac{	an 	heta}{1-\cot 	heta}+\frac{\cot 	heta}{1-	an 	heta}=1+\sec 	heta \operatorname{cosec} 	heta$

$L.H.S.$ $=\frac{	an 	heta}{1-\co

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Vedclass · Answer

$\frac{	an 	heta}{1-\cot 	heta}+\frac{\cot 	heta}{1-	an 	heta}=1+\sec 	heta \operatorname{cosec} 	heta$

$L.H.S.$ $=\frac{	an 	heta}{1-\cot 	heta}+\frac{\cot 	heta}{1-	an 	heta}$

$=\frac{\frac{\sin 	heta}{\cos 	heta}}{1-\frac{\cos 	heta}{\sin 	heta}}+\frac{\frac{\cos 	heta}{\sin 	heta}}{1-\frac{\sin 	heta}{\cos 	heta}}$

$=\frac{\frac{\sin 	heta}{\cos 	heta}}{\frac{\sin 	heta-\cos 	heta}{\sin 	heta}}+\frac{\frac{\cos 	heta}{\sin 	heta}}{\frac{\cos 	heta-\sin 	heta}{\cos 	heta}}$

$=\frac{\sin ^{2} 	heta}{\cos 	heta(\sin 	heta-\cos 	heta)}-\frac{\cos ^{2} 	heta}{\sin 	heta(\sin 	heta-\cos 	heta)}$

$=\frac{1}{(\sin 	heta-\cos 	heta)}\left[\frac{\sin ^{2} 	heta}{\cos 	heta}-\frac{\cos ^{2} 	heta}{\sin 	heta}ight]$

$=\left(\frac{1}{\sin 	heta-\cos 	heta}ight)\left[\frac{\sin ^{3} 	heta-\cos ^{3} 	heta}{\sin 	heta \cos 	heta}ight]$

$=\left(\frac{1}{\sin 	heta-\cos 	heta}ight)\left[\frac{(\sin 	heta-\cos 	heta)\left(\sin ^{2} 	heta+\cos ^{2} 	heta+\sin 	heta \cos 	hetaight)}{\sin 	heta \cos 	heta}ight]$

$=\frac{(1+\sin 	heta \cos 	heta)}{(\sin 	heta \cos 	heta)}$

$=\sec 	heta \operatorname{cosec} 	heta+1$

$= R . H.S.$

Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta$

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