Question

Prove the following :
$\frac{\tan \theta }{sec\theta +1}+\frac{sec\theta +1}{\tan \theta }=2\cos ec\theta$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\frac{\tan \theta }{sec\theta +1}+\frac{sec\theta +1}{\tan \theta } \\ =\frac{{\displaystyle \frac{\sin \theta }{\cos \theta }}}{{\displaystyle \frac{1}{\cos \theta }}+1}+\frac{{\displaystyle \frac{1}{\cos \theta }}+1}{{\displaystyle \frac{\sin \theta }{\cos \theta }}} \\ =\frac{{\displaystyle \frac{\sin \theta }{\cos \theta }}}{{\displaystyle \frac{1+\cos \theta }{\cos \theta }}}+\frac{{\displaystyle \frac{1+\cos \theta }{\cos \theta }}}{{\displaystyle \frac{\sin \theta }{\cos \theta }}} \\ =\frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta } \\ =\frac{{\sin }^2\theta +(1+\cos \theta {)}^2}{(1+\cos \theta )\sin \theta } \\ =\frac{{\sin }^2\theta +1+2\cos \theta +{\cos }^2\theta }{(1+\cos \theta )\sin \theta } \\ =\frac{1+1+2\cos \theta }{(1+\cos \theta )\sin \theta } \\ =\frac{2+2\cos \theta }{(1+\cos \theta )\sin \theta } \\ =\frac{2(1+\cos \theta )}{(1+\cos \theta )\sin \theta } \\ =\frac{2}{\sin \theta } \\ =2\cos ec\theta \\ =\mathrm{RHS} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$