Question

Prove that $\mathrm{secA}-tanA=\frac{\cos A}{1+\sin A}$

Answer 1

L.H.S

$\sec A-\tan A$

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

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

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

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

$=\frac{{\cos }^{2}A}{\cos A(1+\sin A)}[\because 1-{\sin }^{2}A={\cos }^{2}A]$

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

Hence, proved.