Given:
We are required to prove:
[tex]\frac{\sec\text{ }\theta\text{ }}{\tan\text{ }\theta\text{ + cot}\theta}\text{ = sin}\theta[/tex]
From the left-hand side:
[tex]\begin{gathered} =\frac{\sec\text{ }\theta\text{ }}{\tan\text{ }\theta\text{ + cot}\theta}\text{ } \\ =\text{ }\frac{\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}\text{ + }\frac{\cos \theta}{\sin \theta}} \\ =\text{ }\frac{\frac{1}{\cos\theta}}{\frac{\sin ^2\theta+cos^2\theta}{\sin \theta\cos \theta}} \\ \end{gathered}[/tex]
From standard trigonometric identity, we have:
[tex]\sin ^2\theta+cos^2\theta\text{ = 1}[/tex]
Substituting we have:
[tex]\begin{gathered} =\text{ }\frac{\frac{1}{\cos\theta}}{\frac{1}{\sin \theta\cos \theta}} \\ =\text{ }\frac{\sin \theta\cos \theta}{\cos \theta} \\ =\text{ sin }\theta\text{ (Right-hand side)} \end{gathered}[/tex]
