Question

If sec θ tan θ = m, show that $\left(\frac{m^2-1}{m^2+1}\right)=\mathrm{sin\theta }.$

Answer 1

$\frac{{\mathrm{m}}^2-1}{{\mathrm{m}}^2+1}=\frac{{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)}^2-1}{{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)}^2+1}$
$=\frac{se{c}^2\theta +{\tan }^2\theta +2sec\theta \tan \theta -1}{se{c}^2\theta +{\tan }^2\theta +2sec\theta \tan \theta +1}$

$=\frac{2{\tan }^2\theta +2sec\theta \tan \theta }{2se{c}^2\theta +2sec\theta \tan \theta }$ [Since, sec²θ-1=tan²θ and tan²θ+1=sec²θ ]

$$\begin{gathered} =\frac{2\tan \theta \left(\tan \theta +sec\theta \right)}{2sec\theta \left(\tan \theta +sec\theta \right)} \\ =\frac{\tan \theta }{sec\theta } \\ =\left(\frac{\sin \theta }{\cos \theta }\times \cos \theta \right) \\ =\sin \theta \end{gathered}$$
Hence, $\frac{{\mathrm{m}}^2-1}{{\mathrm{m}}^2+1}=\mathrm{sin\theta }$