Question

$\frac{(1+\tan hx)}{(1-\tan hx)}$ is equal to

• A. $e^{2x}$
• B. $e^{-2x}$
• C. $i$
• D. $-1$

Answer 1

The correct option is A

$e^{2x}$

Explanation for the correct option:

Simplify using standard identities

Here we can write, $\frac{\tan hx+1}{\tan hx-1}=\frac{{\displaystyle \frac{\sin hx}{\cos hx}}+1}{{\displaystyle \frac{\sin hx}{\cos hx}}-1}$ $\left[\mathbf{\because }\mathit{t}\mathit{a}\mathit{n}\mathit{h}\mathit{x}\mathbf{=}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{h}\mathbf{x}}{\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{h}\mathbf{x}}\right]$

$=\frac{\sin hx+\cos hx}{\sin hx-\cos hx}$

$=\frac{{\displaystyle \frac{e^x-e^{-x}}{2}}+{\displaystyle \frac{e^x+e^{-x}}{2}}}{\frac{e^x-e^{-x}}{2}-\frac{e^x+e^{-x}}{2}}$ $\left[\mathbf{\because }\mathit{s}\mathit{i}\mathit{n}\mathit{h}\mathit{x}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{x}}\mathbf{-}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}}{\mathbf{2}}\mathbf{,}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{h}\mathit{x}\mathbf{=}\frac{{\mathbf{e}}^{\mathbf{x}}\mathbf{+}{\mathbf{e}}^{\mathbf{-}\mathbf{x}}}{\mathbf{2}}\right]$

$=\frac{e^x}{e^{-x}}={\mathit{e}}^{\mathbf{2}\mathbf{x}}$

Hence, Option ‘A’ is Correct.