Question

Prove that:
$\frac{\tan \theta +\sin \theta }{\tan \theta -\sin \theta }=\frac{\sec \theta +1}{\sec \theta -1}$

Answer 1

$\begin{array}{c}\frac{\tan \theta +\sin \theta }{\tan \theta -\sin \theta }=\frac{\sec \theta +1}{\sec \theta -1}\\ L.H.S=\frac{\tan \theta +\sin \theta }{\tan \theta -\sin \theta }\\ =\frac{\frac{\sin \theta }{\cos \theta }+\frac{\sin \theta }{1}}{\frac{\sin \theta }{\cos \theta }-\frac{\sin \theta }{1}}\\ =\frac{\frac{\sin \theta +\sin \theta .\cos \theta }{\cos \theta }}{\frac{\sin \theta -\sin \theta .\cos \theta }{\cos \theta }}\\ =\frac{\sin \theta +\sin \theta .\cos \theta }{\sin \theta -\sin \theta .\cos \theta }\\ =\frac{\sin \theta \left(1+\cos \theta \right)}{\sin \theta \left(1-\cos \theta \right)}\\ =\frac{1+\frac{1}{\sec \theta }}{1-\frac{1}{\sec \theta }}\\ =\frac{\sec \theta +1}{\sec \theta -1}\\ L.H.S=R.H.SProved.\\ R.H.S=\frac{\tan \theta +\sin \theta }{\tan \theta -\sin \theta }\\ =\frac{\sin \theta \left(\frac{\tan \theta }{\sin \theta }+1\right)}{\sin \theta \left(\frac{\tan \theta }{\sin \theta }-1\right)}\\ =\frac{\frac{\sin \theta }{\cos \theta .\sin \theta }+1}{\frac{\sin \theta }{\cos \theta .\sin \theta }-1}\\ =\frac{\frac{1}{\cos \theta }+1}{\frac{1}{\cos \theta }-1}\\ =\frac{\sec \theta +1}{\sec \theta -1}\\ L.H.S=R.H.Sproved.\end{array}$