Question

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

Answer 1

$\frac{{\mathrm{m}}^2-1}{{\mathrm{m}}^2+1}=\frac{{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)}^2-1}{{\left(\mathrm{sec\theta }+\mathrm{tan\theta }\right)}^2+1}$
$=\frac{{\sec }^2\mathrm{\theta }+{\tan }^2\mathrm{\theta }+2\mathrm{sec\theta tan\theta }-1}{{\sec }^2\mathrm{\theta }+{\tan }^2\mathrm{\theta }+2\mathrm{sec\theta tan\theta }+1}$ [Since sec²θ - 1 = tan²θ and tan²θ + 1 = sec²θ]
$=\frac{2{\tan }^2\mathrm{\theta }+2\mathrm{sec\theta tan\theta }}{2{\sec }^2\mathrm{\theta }+2\mathrm{sec\theta tan\theta }}$
$=\frac{2\mathrm{tan\theta }\left(\mathrm{tan\theta }+\mathrm{sec\theta }\right)}{2\mathrm{sec\theta }\left(\mathrm{tan\theta }+\mathrm{sec\theta }\right)}=\frac{\mathrm{tan\theta }}{\mathrm{sec\theta }}$
$=\left(\frac{\mathrm{sin\theta }}{\mathrm{cos\theta }}\times \cos \mathrm{\theta }\right)=\sin \mathrm{\theta }$
Hence, $\frac{{\mathrm{m}}^2-1}{{\mathrm{m}}^2+1}=\mathrm{sin\theta }$