Question

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

Answer 1

Simplifying the LHS of $\frac{\tan \theta }{\sec \theta -1}=\frac{\tan \theta +\sec \theta +1}{\tan \theta +\sec \theta -1}$.

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

$=\frac{\frac{\sin \theta }{\cos \theta }}{\frac{1-\cos \theta }{\cos \theta }}$

$=\frac{\sin \theta }{1-\cos \theta }$

$=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}}{2{\sin }^2\frac{\theta }{2}}$

$=\cot \frac{\theta }{2}$

Simplifying the RHS of $\frac{\tan \theta }{\sec \theta -1}=\frac{\tan \theta +\sec \theta +1}{\tan \theta +\sec \theta -1}$

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

$=\frac{\sin \theta +\cos \theta +1}{\sin \theta -\cos \theta +1}$

$=\frac{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}+2{\cos }^2\frac{\theta }{2}}{2\sin \frac{\theta }{2}\cos \frac{\theta }{2}+2{\sin }^2\frac{\theta }{2}}$

$=\frac{2\cos \frac{\theta }{2}\left(\sin \frac{\theta }{2}+\cos \frac{\theta }{2}\right)}{2\sin \frac{\theta }{2}\left(\cos \frac{\theta }{2}+\sin \frac{\theta }{2}\right)}$

$=\cot \frac{\theta }{2}$

It can be observed that $LHS=RHS$.

Hence proved.