Question

If $\sin \theta =\frac{a}{b}$, show that $(\sec \theta +\tan \theta )=\sqrt{\frac{mb+a}{mb-a}}.$.Find $m$

Answer 1

$\sin \theta =\frac{a}{b}=\frac{P}{H}$ (Refer to Image)

$B=\sqrt{b^2-a^2}$

$\sec \theta =\frac{1}{\cos \theta }=\frac{H}{B}=\frac{b}{\sqrt{b^2-a^2}}$

$\tan \theta =\frac{\sin \theta }{\cos \theta }=\frac{\frac{a}{b}}{\frac{\sqrt{b^2-a^2}}{b}}=\frac{a}{\sqrt{b^2-a^2}}$

$\Rightarrow (\sec \theta +\tan \theta )=\frac{b}{\sqrt{b^2-a^2}}+\frac{a}{\sqrt{b^2-a^2}}$

$=\frac{b+a}{\sqrt{b^2-a^2}}$

$=\sqrt{\frac{(b+a{)}^2}{b^2-a^2}}$

$=\sqrt{\frac{(b+a{)}^2}{(b-a)(b+a)}}$

$\Rightarrow \sqrt{\frac{b+a}{b-a}}=\sqrt{\frac{mb+a}{mb-a}}$

$\therefore m=1$