Question

If sec θ + tan θ = x, prove that sin θ = $\frac{{\mathit{x}}^{\mathit{2}}\mathit{-}\mathit{1}}{{\mathit{x}}^{\mathit{2}}\mathit{+}\mathit{1}}\mathit{.}$

Answer 1

We have,
$\frac{x^2-1}{x^2+1}=\frac{{\left(sec\theta +\tan \theta \right)}^2-1}{{\left(sec\theta +\tan \theta \right)}^2+1}$

$=\frac{se{c}^2\theta +{\tan }^2\theta +2sec\theta \tan \theta -1}{se{c}^2\theta +{\tan }^2\theta +2sec\theta \tan \theta +1}$

$=\frac{2{\tan }^2\theta +2sec\theta \tan \theta }{2se{c}^2\theta +2sec\theta \tan \theta }\left[\mathrm{since}\mathrm{s}\mathrm{e}\mathrm{c}\right]$²θ-1=tan²θ

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

Hence, $\frac{x^2-1}{x^2+1}=\sin \theta$