Question

${\displaystyle \sqrt{\frac{1+\cos \theta }{1-\cos \theta }}+\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}=2\cos ec\theta }$

 

Answer 1

LHS=${\displaystyle \sqrt{\frac{1+\cos \theta }{1-\cos \theta }}+\sqrt{\frac{1-\cos \theta }{1+\cos \theta }}}$

    =${\displaystyle \sqrt{\frac{{(1+\cos \theta )}^2}{(1-\cos \theta )(1+\cos \theta )}}+\sqrt{\frac{{(1-\cos \theta )}^2}{(1+\cos \theta )(1-\cos \theta )}}}$

    =${\displaystyle \sqrt{\frac{{(1+\cos \theta )}^2}{(1-{\cos }^2\theta )}}+\sqrt{\frac{{(1-\cos \theta )}^2}{(1-{\cos }^2\theta )}}}$

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

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

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

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

    = 2cos ecθ
   = RHS