Question

If $\cos (\alpha -\beta )=1$ and $\cos (\alpha +\beta )=\frac{1}{e}$, where $-\pi <\alpha ,\beta <\pi$, then total number of ordered pair of $(\alpha ,\beta )$ is

• A. $0$
• B. $1$
• C. $2$
• D. $4$

Answer 1

Answer: (D)

$4$

Given, $\cos (\alpha -\beta )=1$ and $\cos (\alpha +\beta )=\frac{1}{e}$

We have $-\pi <\alpha ,\beta <\pi$

$-2\pi <\alpha +\beta <2\pi$

$\Rightarrow -2\pi <\alpha -\beta <2\pi$
$\cos (\alpha -\beta )=1\Rightarrow \alpha -\beta =0\Rightarrow \alpha =\beta$

Also, $\cos (\alpha +\beta )=\frac{1}{e}$, then $\cos 2\alpha =\frac{1}{e}$
$\Rightarrow -2\pi <2\alpha <2\pi$
Hence, the were will be four solutions.