Question

Prove that $\frac{1+\cos \theta }{1-\cos \theta }=\frac{{\tan }^2\theta }{{\left(\sec \theta -1\right)}^2}$

Answer 1

$R.H.S.$

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

Divide and multiply above expression by $1-\cos \theta$ we get

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

$(\frac{1-{\cos }^2\theta }{(1-\cos \theta {)}^2})$ $\because (a+b)(a-b)=(a^2-b^2)$

$(\frac{{\sin }^2\theta }{(1-\cos \theta {)}^2})$

divide numerator and denominator by $co{s}^2\theta$ we get

$(\frac{\frac{{\sin }^2\theta }{{\cos }^2\theta }}{\frac{(1-\cos \theta {)}^2}{{\cos }^2\theta }})$

$\frac{ta{n}^2\theta }{(\frac{1}{\cos \theta }-1{)}^2}$ $\because \frac{\sin \theta }{\cos \theta }=\tan \theta$

$\frac{{\tan }^2\theta }{(\sec \theta -1{)}^2}=L.H.S$

Hence proved.