Question

Prove the following :

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

Answer 1

$$\begin{gathered} \mathrm{RHS}=\frac{\tan A+secA+1}{\tan A+secA-1} \\ =\frac{\tan A+(1+secA)}{\tan A-(1-secA)} \\ =\frac{\tan A+(1+secA)}{\tan A-(1-secA)}\times \frac{\tan A+(1-secA)}{\tan A+(1-secA)} \\ =\frac{\tan {}^{2}A+\tan A(1-secA)+\tan A(1+secA)+(1+secA)(1-secA)}{\tan {}^{2}A-{(1-secA)}^2} \\ =\frac{\tan {}^{2}A+\tan A-\tan AsecA+\tan A+\tan AsecA+1-se{c}^{2}A}{\tan {}^{2}A-(1+se{c}^{2}A-2secA)} \\ =\frac{1+\tan {}^{2}A+2\tan A-se{c}^{2}A}{\tan {}^{2}A-1-se{c}^{2}A+2secA} \\ =\frac{2\tan A}{-1-1+2secA} \\ =\frac{2\tan A}{2(secA-1)} \end{gathered}$$

$$\begin{gathered} =\frac{\tan A}{secA-1} \\ =\mathrm{LHS} \\ \mathrm{Hence},\mathrm{proved}. \end{gathered}$$