Question

If $\sin \theta =\frac{a}{b}$, then find the value of $\sec \theta +\tan \theta$

Answer 1

$sin\theta =\frac{a}{b}cos\theta =\sqrt{1-si{n}^2\theta }$

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

$sec\theta +tan\theta$

$\frac{1}{cos\theta }+\frac{sin\theta }{cos\theta }=\frac{1+sin\theta }{cos\theta }$

$\frac{1+\frac{a}{b}}{\sqrt{\frac{b^{2-a^2}}{b}}}=\frac{b+a}{\sqrt{b^2{a}^2}}$

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