TS EAMCET · Maths · Inverse Trigonometric Functions
\(\frac{1+\tanh \frac{x}{2}}{1-\tanh \frac{x}{2}}\) is equal to
- A \(e^{-x}\)
- B \(e^{x}\)
- C \(2 e^{x / 2}\)
- D \(2 e^{-x / 2}\)
Answer & Solution
Correct Answer
(B) \(e^{x}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \frac{1+\tanh \frac{x}{2}}{1-\tanh \frac{x}{2}} & =\frac{\cosh \frac{x}{2}+\sinh \frac{x}{2}}{\cosh \frac{x}{2}-\sinh \frac{x}{2}} \\ & =\frac{e^{x / 2}}{e^{-x / 2}}=e^x\end{aligned}\)
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