THE AREA OF THE TRIANGLE FORMED BY THE AXES AND THE LINE (cos h alpha - sin h alpha)x + (cos h alpha + sin h alpha)y = 2 in square units, is
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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ \left\{\cosh \left(\mathrm{\alpha }\right)-\sinh \left(\mathrm{\alpha }\right)\right\}\mathrm{x}+\left\{\cosh \left(\mathrm{\alpha }\right)+\sinh \left(\mathrm{\alpha }\right)\right\}\mathrm{y}=2 \\ \mathrm{For} \mathrm{X}-\mathrm{axis} \mathrm{intercept}, \mathrm{put} \mathrm{y}=0 \\ \Rightarrow \left\{\cosh \left(\mathrm{\alpha }\right)-\sinh \left(\mathrm{\alpha }\right)\right\}\mathrm{x}=2 \\ \Rightarrow \mathrm{x}=\frac{2}{\cosh \left(\mathrm{\alpha }\right)-\sinh \left(\mathrm{\alpha }\right)} \\ \mathrm{Length} \mathrm{of} \mathrm{X}-\mathrm{axis} \mathrm{intercept}\left({\mathrm{l}}_{\mathrm{x}}\right)=\left|\mathrm{x}\right|=\frac{2}{\left|\cosh \left(\mathrm{\alpha }\right)-\sinh \left(\mathrm{\alpha }\right)\right|} \\ \mathrm{For} \mathrm{Y}-\mathrm{axis} \mathrm{intercept}, \mathrm{put} \mathrm{x}=0 \\ \left\{\cosh \left(\mathrm{\alpha }\right)+\sinh \left(\mathrm{\alpha }\right)\right\}\mathrm{y}=2 \\ \Rightarrow \mathrm{y}=\frac{2}{\left|\cosh \left(\mathrm{\alpha }\right)+\sinh \left(\mathrm{\alpha }\right)\right|} \\ \mathrm{Length} \mathrm{of} \mathrm{Y}-\mathrm{axis} \mathrm{intercept}\left({\mathrm{l}}_{\mathrm{y}}\right)=\left|\mathrm{y}\right|=\frac{2}{\left|\cosh \left(\mathrm{\alpha }\right)+\sinh \left(\mathrm{\alpha }\right)\right|} \\ \mathrm{Area} \mathrm{of} \mathrm{Tringale}\left(\mathrm{A}\right) =\frac{1}{2}\times {\mathrm{l}}_{\mathrm{x}}\times {\mathrm{l}}_{\mathrm{y}} \\ =\frac{1}{2}\times \frac{2}{\left|\cosh \left(\mathrm{\alpha }\right)-\sinh \left(\mathrm{\alpha }\right)\right|}\times \frac{2}{\left|\cosh \left(\mathrm{\alpha }\right)+\sinh \left(\mathrm{\alpha }\right)\right|} \\ =\frac{2}{\left|\left\{\cosh \left(\mathrm{\alpha }\right)-\sinh \left(\mathrm{\alpha }\right)\right\}.\left\{\cosh \left(\mathrm{\alpha }\right)+\sinh \left(\mathrm{\alpha }\right)\right\}\right|} \\ \mathrm{A}=\frac{2}{\left|{\cosh }^2\left(\mathrm{\alpha }\right)-{\sinh }^2\left(\mathrm{\alpha }\right)\right|} ;\mathrm{equation}\left(\mathrm{i}\right) \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ \cosh \left(\mathrm{\alpha }\right)=\frac{{\mathrm{e}}^{\mathrm{\alpha }}+{\mathrm{e}}^{-\mathrm{\alpha }}}{2} \mathrm{and} \sinh \left(\mathrm{\alpha }\right)=\frac{{\mathrm{e}}^{\mathrm{\alpha }}-{\mathrm{e}}^{-\mathrm{\alpha }}}{2} \\ {\cosh }^2\left(\mathrm{\alpha }\right)-{\sinh }^2\left(\mathrm{\alpha }\right)={\left(\frac{{\mathrm{e}}^{\mathrm{\alpha }}+{\mathrm{e}}^{-\mathrm{\alpha }}}{2}\right)}^2-{\left(\frac{{\mathrm{e}}^{\mathrm{\alpha }}-{\mathrm{e}}^{-\mathrm{\alpha }}}{2}\right)}^2 \\ =\frac{1}{4}\left({\mathrm{e}}^{2\mathrm{\alpha }}+{\mathrm{e}}^{-2\mathrm{\alpha }}+2{\mathrm{e}}^{\mathrm{\alpha }}{\mathrm{e}}^{-\mathrm{\alpha }}-{\mathrm{e}}^{2\mathrm{\alpha }}-{\mathrm{e}}^{-2\mathrm{\alpha }}+2{\mathrm{e}}^{\mathrm{\alpha }}{\mathrm{e}}^{-\mathrm{\alpha }}\right) \\ =\frac{1}{4}\left(2+2\right) \\ =\frac{1}{4}\times 4=1 \\ \Rightarrow {\cosh }^2\left(\mathrm{\alpha }\right)-{\sinh }^2\left(\mathrm{\alpha }\right)=1 \\ \mathrm{Putting} \mathrm{this} \mathrm{value} \mathrm{in} \mathrm{equation}\left(\mathrm{i}\right), \mathrm{we} \mathrm{get}, \\ \mathrm{A}=\frac{2}{\left|1\right|}=2 {\mathrm{unit}}^2 \\ \mathbf{Hence}\mathbf{ }\mathbf{option}\left(\mathbf{3}\right)\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{correct}\mathbf{.} \end{gathered}$$

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