If $\sin \theta=\frac{a}{b}$, show that $(\sec \theta+\tan \theta)=\sqrt{\frac{b+a}{b-a}}$.
$\mathrm{LHS}=(\sec \theta+\tan \theta)$
$=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}$
$=\frac{1+\sin \theta}{\cos \theta}$
$=\frac{1+\sin \theta}{\sqrt{1-\sin ^{2} \theta}}$
$=\frac{\left(1+\frac{a}{b}\right)}{\sqrt{1-\left(\frac{a}{b}\right)^{2}}}$
$=\frac{\left(\frac{1}{1}+\frac{a}{b}\right)}{\sqrt{\frac{1}{1}-\frac{a^{2}}{b^{2}}}}$
$=\frac{\left(\frac{b+a}{b}\right)}{\sqrt{\frac{b^{2}-a^{2}}{b^{2}}}}$
$=\frac{\left(\frac{b+a}{b}\right)}{\left(\frac{\sqrt{b^{2}-a^{2}}}{b}\right)}$
$=\frac{(b+a)}{\sqrt{(b+a)(b-a)}}$
$=\frac{(b+a)}{\sqrt{(b+a)} \sqrt{(b-a)}}$
$=\frac{\sqrt{(b+a)}}{\sqrt{(b-a)}}$
$=\sqrt{\frac{b+a}{b-a}}$
$=\mathrm{RHS}$
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