Question

Solve:
$\frac{1}{\sec \theta -\tan \theta }-\frac{1}{\cos \theta }=\frac{1}{\cos \theta }-\frac{1}{\sec \theta +\tan \theta }$

Answer 1

$\therefore \frac{1}{\sec \theta -\tan \theta }-\frac{1}{\cos \theta }=\frac{1}{\cos \theta }-\frac{1}{\sec \theta +\tan \theta }$

$\Rightarrow$ Solving LHS

$\Rightarrow \frac{1}{\sec \theta -\tan \theta }-\frac{1}{\cos \theta }=\frac{\cos \theta -\sec \theta +\tan \theta }{(\sec \theta -\tan \theta )\cos \theta }$

$\Rightarrow \frac{\cos \theta -\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta }}{(\frac{1}{\cos \theta }-\frac{\sin \theta }{\cos \theta })\cos \theta }=\frac{\frac{{\cos }^2\theta +\sin \theta -1}{\cos \theta }}{\frac{(1-\sin \theta )}{\cos \theta }.\cos \theta }$

$\Rightarrow \frac{{\cos }^2\theta +\sin \theta -1}{\cos \theta [1-\sin \theta ]}=\frac{{\cos }^2\theta -\sin \theta -{\cos }^2\theta -{\sin }^2\theta }{\cos \theta [1-\sin \theta ]}$

$\Rightarrow \frac{\sin \theta }{\cos \theta }\frac{[1-\sin \theta ]}{1-\sin \theta ]}=\tan \theta =$ LHS

now solving RHS

$\Rightarrow \frac{1}{\cos \theta }-\frac{1}{(\sec \theta +\tan \theta )}=\frac{\sec \theta +\tan \theta -\cos \theta }{\cos \theta \left(\frac{1+\sin \theta }{\cos \theta }\right)}$

$\Rightarrow \frac{1+\sin \theta -{\cos }^2\theta }{(1+\sin \theta )\cos \theta }\Rightarrow \frac{\sin \theta [1+\sin \theta ]}{[1+\sin \theta ]\cos \theta }$

$\Rightarrow \tan \theta =$ RHS

$\therefore$

[LHS=RHS=$\tan \theta$].