Question

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

Answer 1

$$\begin{gathered} \mathrm{LHS}=\left(\sec \theta +\tan \theta \right) \\ =\frac{1}{\cos \theta }+\frac{\sin \theta }{\cos \theta } \\ =\frac{1+\sin \theta }{\cos \theta } \\ =\frac{1+\sin \theta }{\sqrt{1-{\sin }^2\theta }} \\ =\frac{\left(1+{\displaystyle \frac{a}{b}}\right)}{\sqrt{1-{\left({\displaystyle \frac{a}{b}}\right)}^2}} \end{gathered}$$
$$\begin{gathered} =\frac{\left({\displaystyle \frac{1}{1}+\frac{a}{b}}\right)}{\sqrt{{\displaystyle \frac{1}{1}-\frac{a^2}{b^2}}}} \\ =\frac{\left({\displaystyle \frac{b+a}{b}}\right)}{\sqrt{{\displaystyle \frac{b^2-a^2}{b^2}}}} \\ =\frac{\left({\displaystyle \frac{b+a}{b}}\right)}{\left({\displaystyle \frac{\sqrt{b^2-a^2}}{b}}\right)} \\ =\frac{\left({\displaystyle b+a}\right)}{{\displaystyle \sqrt{\left(b+a\right)\left(b-a\right)}}} \end{gathered}$$
$$\begin{gathered} =\frac{\left({\displaystyle b+a}\right)}{{\displaystyle \sqrt{\left(b+a\right)}\sqrt{\left(b-a\right)}}} \\ =\frac{\sqrt{\left(b+a\right)}}{{\displaystyle \sqrt{\left(b-a\right)}}} \\ =\sqrt{\frac{b+a}{b-a}} \\ =\mathrm{RHS} \end{gathered}$$