Question

${\displaystyle \frac{\tan \theta }{(\sec \theta -1)}+\frac{\tan \theta }{(\sec \theta +1)}=2\sec \theta }$

Answer 1

LHS = ${\displaystyle \frac{\tan \theta }{(\sec \theta -1)}+\frac{\tan \theta }{(\sec \theta +1)}}$

      =${\displaystyle \tan \theta \left\{\frac{{\sec }^{\theta }+1+\sec \theta -1}{(\sec \theta -1)(\sec \theta +1)}\right\}}$

       =${\displaystyle \tan \theta \left\{\frac{2\sec \theta }{{\sec }^2\theta -1}\right\}}$

      =${\displaystyle \tan \theta \times \frac{2\sec \theta }{{\tan }^2\theta -1}}$

      =${\displaystyle 2\frac{{\sec }^{\theta }}{{\tan }^{\theta }}}$

     =${\displaystyle 2\frac{\frac{1}{\cos \theta }}{\frac{\sin \theta }{\cos \theta }}}$

     =${\displaystyle 2\frac{1}{\sin \theta }}$

    =${\displaystyle 2\cos ec \theta }$

    = RHS
Hence, LHS = RHS