Question

If $\sec \theta +\tan \theta =m(>1),$ then the value of $\sin \theta$ is $(0^o<\theta <{90}^o)$

Answer 1

$\begin{array}{c}\sec \theta +\tan \theta =m\\ \sec \theta -\tan \theta =\frac{1}{m}\\ 2\sec \theta =\left(m+\frac{1}{m}\right)\\ \sec \theta =\left(\frac{m^2+1}{2m}\right)\\ \cos \theta =\frac{2m}{m^2+1}\\ \sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-\frac{4m^2}{{\left(m^2+1\right)}^2}}\\ =\frac{\sqrt{m^4+1-2m^2}}{m^2+1}\\ =\frac{\left(m^2-1\right)}{m^2+1}\\ =\left(\frac{m^2-1}{m^2+1}\right)\\ Hence,whichistherequiredanswer.\end{array}$