Question

Prove that: $\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}$.

Answer 1

$L.H.S=\frac{\tan A+\sec A-1}{\tan A-\sec A+1}$

$\begin{array}{c}=\frac{\left(\tan A+\sec A\right)-\left({\sec }^{2}A-{\tan }^{2}A\right)}{\tan A-\sec A+1}\\ =\frac{\left(\tan A+\sec A\right)\left(1-\left(\sec A-\tan A\right)\right)}{\tan A-\sec A+1}\\ =\frac{\left(\tan A+\sec A\right)\left(1-\sec A-\tan A\right)}{\tan A+1-\sec A}\end{array}$

$=\sec A+\tan A$

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

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

$R.H.S$ proved.