Question

Prove that \(\displaystyle \frac{\sin A-\cos A+1}{\sin A+\cos A-1}=\frac{1}{\sec A-\tan A}\).

Hint:

When a proof involves $\sec A\pm\tan A$, convert to $\dfrac{1\pm\sin A}{\cos A}$ and try multiplying the other side by $(1\mp\sin A)$ to reveal a common factor.

Answer 1

Start with the LHS. Multiply numerator and denominator by $(1-\sin A)$: \[ \frac{\sin A-\cos A+1}{\sin A+\cos A-1} =\frac{(\sin A-\cos A+1)(1-\sin A)}{(\sin A+\cos A-1)(1-\sin A)}. \] Expand the numerator and use $1-\sin^2 A=\cos^2 A$: \[ (\sin A-\cos A+1)(1-\sin A)=1-\sin^2 A-\cos A+\sin A\cos A =\cos^2 A-\cos A+\sin A\cos A =\cos A\big(\cos A-1+\sin A\big). \] Note that $\cos A-1+\sin A=(\sin A+\cos A-1)$. Hence \[ \frac{\sin A-\cos A+1}{\sin A+\cos A-1} =\frac{\cos A(\sin A+\cos A-1)}{(\sin A+\cos A-1)(1-\sin A)} =\frac{\cos A}{1-\sin A}. \] But \[ \frac{\cos A}{1-\sin A}=\frac{1}{\dfrac{1-\sin A}{\cos A}} =\frac{1}{\sec A-\tan A}. \] Therefore, \[ \boxed{\displaystyle \frac{\sin A-\cos A+1}{\sin A+\cos A-1}=\frac{1}{\sec A-\tan A}}. \]