Question

If $\frac{\cos x}{\cos y}=2$ and $\cos (x-y)=\frac{\sqrt{3}}{2}$ then $y=$

• A. ${\tan }^{-1}\left(2.266\right)$
• B. ${\tan }^{-1}(-2.266)$
• C. $\text{both a and b}$
• D. $\text{none}$

Answer 1

Answer: (C)

$\text{both a and b}$

$\cos (x-y)=\frac{\sqrt{3}}{2}$

$x-y=\frac{\pi }{6},\frac{-\pi }{6}$

$\frac{\cos x}{\cos y}=2$

$\cos x=2\cos y$

when, $x=y+\frac{\pi }{6}$

$\cos x=\cos (y+\frac{\pi }{6})=2\cos y$

$\frac{\sqrt{3}}{2}\cos y-\frac{1}{2}\sin y=2\cos y$

$(\frac{\sqrt{3}}{2}-2)\cos y=\frac{1}{2}\sin y$

$-1.134\times \cos y=\frac{1}{2}\sin y$

$\tan y=-2.268$

$y={\tan }^{-1}(-2.268)$

when $x=y-\frac{\pi }{6}$

$\cos (y-\frac{\pi }{6})=2\cos y$

$\frac{\sqrt{3}}{2}\cos y+\frac{1}{2}\sin y=2\cos y$

$\tan y=2.268$

$y={\tan }^{-1}\left(2.268\right)$

$\left(a\right)$ and $\left(b\right)$ both.