If frac{sec theta+tan theta}{sec theta-tan th | vedclass

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Question

(A) Given: $\frac{\sec 	heta+	an 	heta}{\sec 	heta-	an 	heta} = 2 \frac{51}{79} = \frac{2 	imes 79 + 51}{79} = \frac{158 + 51}{79} = \frac{209}

Vedclass · Accepted Answer

$\frac{65}{144}$

Vedclass · Answer

(A) Given: $\frac{\sec 	heta+	an 	heta}{\sec 	heta-	an 	heta} = 2 \frac{51}{79} = \frac{2 	imes 79 + 51}{79} = \frac{158 + 51}{79} = \frac{209}{79}$.Apply Componendo and Dividendo $(C \& D)$:$\frac{(\sec 	heta+	an 	heta) + (\sec 	heta-	an 	heta)}{(\sec 	heta+	an 	heta) - (\sec 	heta-	an 	heta)} = \frac{209 + 79}{209 - 79}$.$\frac{2 \sec 	heta}{2 	an 	heta} = \frac{288}{130}$.$\frac{1}{\cos 	heta} 	imes \frac{\cos 	heta}{\sin 	heta} = \frac{144}{65}$.$\frac{1}{\sin 	heta} = \frac{144}{65}$.Therefore,$\sin 	heta = \frac{65}{144}$.

If $\frac{\sec \theta+\tan \theta}{\sec \theta-\tan \theta}=2 \frac{51}{79}$,then the value of $\sin \theta$ is

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