If $\tan \mathrm{h} x=\operatorname{sech} y=\frac{3}{5}$ and $e^{\mathrm{x}+\mathrm{y}}$ is an integer, then…

If $\tan \mathrm{h} x=\operatorname{sech} y=\frac{3}{5}$ and $e^{\mathrm{x}+\mathrm{y}}$ is an integer, then $e^{\mathrm{x}+\mathrm{y}}=$
  1. 2
  2. 8
  3. 1
  4. 6

Solution

$\begin{aligned} & \tan h x=\frac{3}{5} \Rightarrow \frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{3}{5} \Rightarrow \frac{e^{2 x}-1}{e^{2 x}+1}=\frac{3}{5} \\ & \Rightarrow 5 e^{2 x}-5=3 e^{2 x}+1 \Rightarrow e^{2 x}=4 \Rightarrow e^x=2 \\ & \text { And, } \sec h y=\frac{3}{5} \Rightarrow \frac{2}{e^y+e^{-y}}=\frac{3}{5} \Rightarrow \frac{2 e^y}{e^{2 y}+1}=\frac{3}{5} \\ & \Rightarrow 10 t=3 t^2+3 \\ & \qquad \quad\left[\because \text { Let } e^y=t\right] \\ & \Rightarrow 3 t^2-10 t+3=0 \Rightarrow t=3, \frac{1}{3} \Rightarrow e^y=3, \frac{1}{3} \end{aligned}$
So, $e^{x+y}=e^x \cdot e^y=2 \times 3=6$.

Asked in: AP EAMCET 2024 (22 May Shift 1)

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