Question

Prove that, $\sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A$
Or
If $\tan \theta +\cot \theta =2$, find the value of ${\tan }^3\theta +7{\cot }^3\theta$.

Answer 1

$L.H.S$

$=\sqrt{\frac{1+\sin A}{1-\sin A}}$

$=\sqrt{\frac{1+\sin A}{1-\sin A}\times \frac{1+\sin A}{1+\sin A}}$

$=\sqrt{\frac{(1+\sin A{)}^2}{1-{\sin }^{2}A}}$

$=\sqrt{\frac{(1+\sin A{)}^2}{{\cos }^{2}A}}$

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

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

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

$=\sec A+\tan A=R.H.S\to HenceProved$