Question

$\frac{1+\tan h\left({\displaystyle \frac{x}{2}}\right)}{1-\tan h\left({\displaystyle \frac{x}{2}}\right)}isequalto$

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

Answer 1

The correct option is B

$e^x$

Explanation for the correct option:

Solve the given expression using identities

$\frac{1+\tan h\left({\displaystyle \frac{x}{2}}\right)}{1-\tan h\left({\displaystyle \frac{x}{2}}\right)}$ $=\frac{1+{\displaystyle \frac{\sin h\left({\displaystyle \frac{x}{2}}\right)}{\cos h\left({\displaystyle \frac{x}{2}}\right)}}}{1-\frac{\sin h\left({\displaystyle \frac{x}{2}}\right)}{\cos h\left({\displaystyle \frac{x}{2}}\right)}}$

$=\frac{\sin h{\displaystyle \frac{x}{2}}+\cos h{\displaystyle \frac{x}{2}}}{\cos h{\displaystyle \frac{x}{2}}-\sin h{\displaystyle \frac{x}{2}}}$

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

Hence, Option ‘B’ is Correct.