Question

Prove that $(secA+\tan A-1)(secA-\tan A+1)=2\tan A.$

Answer 1

As we have $(secA+\tan A-1)(secA-\tan A+1)=2\tan A.$

Taking LHS $(secA+\tan A-1)(secA-\tan A+1)$

$$\begin{gathered} \Rightarrow \left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}-1\right)\left(\frac{1}{\cos A}-\frac{\sin A}{\cos A}+1\right) \\ \Rightarrow \left(\frac{1+\sin A-\cos A}{\cos A}\right)\left(\frac{1-\sin A+\cos A}{\cos A}\right) \\ \Rightarrow \frac{1-\sin A+\cos A+\sin A-{\sin }^{2}A+\sin A\cos A-\cos A+\sin A\cos A-{\cos }^{2}A}{{\cos }^{2}A} \\ \Rightarrow \frac{2\sin A\cos A}{{\cos }^{2}A} \\ \Rightarrow 2\tan A \end{gathered}$$