Question

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

Answer 1

$L.H.S=\frac{\sin A-\cos A+1}{\sin A+\cos A-1}$

Dividing by $\cos A$.

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

$=\frac{\tan A-1+secA}{\tan A+1-secA}$………………………($\frac{\sin A}{\cos A}=\tan A,\frac{1}{\cos A}=secA$)

$=\frac{\tan A+secA-1}{\tan A-secA+1}$

$=\frac{\left(secA+\tan A\right)-\left(se{c}^{2}A-{\tan }^{2}A\right)}{\tan A-secA+1}$…………….($se{c}^2\theta -{\tan }^2\theta =1$)

$=\frac{\left(secA+\tan A\right)-\left(secA-\tan A\right)\left(secA+\tan A\right)}{\tan A-secA+1}$

$=\frac{secA+\tan A\left(1-secA-\tan A\right)}{1-secA+\tan A}$

$=secA+\tan A$

$=\frac{1}{secA-\tan A}$.

$=R.H.S.$

Therefore, it is proved that $\frac{\sin A-\cos A+1}{\sin A+\cos A-1}=\frac{1}{secA-\tan A}$