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Trigonometrical Equations
normal
$cos (\alpha \,-\,\beta ) = 1$ and $cos (\alpha +\beta ) = 1/e$ , where $\alpha , \beta \in [-\pi , \pi ]$ . Number of pairs of $(\alpha ,\beta )$ which satisfy both the equations is
A
$0$
B
$1$
C
$2$
D
$4$
Solution
Given that $\operatorname{cost}(\alpha-\beta)=1$ and $\cos (\alpha+\beta)=1 / \mathrm{e}$
where $\alpha, \beta \in[-\pi, \pi]$
Now $\cos (\alpha-\beta)=1 \Rightarrow \alpha-\beta=0$ or $\alpha=\beta$
$\therefore $ $\cos (\alpha+\beta)=1 / e \Rightarrow \cos 2 \alpha=1 / e$
$\because 0<1 / \mathrm{e}<1$ and $2 \alpha \in[-2 \pi, 2 \pi]$
Therefore, there will be four values of $\alpha$ in
$[-2 \pi, 2 \pi]$ and correspondingly four values of $\beta$
Hence, there are four sets of $(\alpha, \beta)$.
Standard 11
Mathematics
