Question

Let $\cos (\alpha -\beta )=1$and $\cos (\alpha +\beta )=\frac{1}{e},$ where $\alpha ,\beta ϵ[-\pi ,\pi ]$ and $e$ is the exponential number.The number of values of the pair $(\alpha ,\beta )$ satisfying both the equations, is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

$\cos (\alpha -\beta )=1$ and $\cos (\alpha +\beta )=\frac{1}{e}$
$\alpha ,\beta ϵ[-\pi ,\pi ]\Rightarrow \alpha +\beta ϵ[-2\pi ,2\pi ]$ and $\alpha -\beta ϵ[-2\pi ,2\pi ]$
thus solution in the given interval are,
$\alpha -\beta =-2\pi ,0,2\pi$, but in this case $\alpha -\beta =-2\pi ,2\pi$ are not possible, $\alpha -\beta =0$ is only possible equation
and $\alpha +\beta =-2\pi +{\cos }^{-1}\left(\frac{1}{e}\right),-{\cos }^{-1}\left(\frac{1}{e}\right),{\cos }^{-1}\left(\frac{1}{e}\right),2\pi -{\cos }^{-1}\left(\frac{1}{e}\right)$
hence solutions are,
$\alpha =\beta =-\pi +\frac{1}{2}{\cos }^{-1}\left(\frac{1}{e}\right),-\frac{1}{2}{\cos }^{-1}\left(\frac{1}{e}\right),\frac{1}{2}{\cos }^{-1}\left(\frac{1}{e}\right),\pi -\frac{1}{2}{\cos }^{-1}\left(\frac{1}{e}\right)$
thus there is $4$ possible pair.