Question

If $\frac{\sin x}{\sin y}=\frac{1}{2}$ and $\frac{\cos x}{\cos y}=\frac{3}{2}$ , where $x,yϵ(0,\frac{\pi }{2})$ then $\sqrt{{\sec }^{2}y-8}=$

• A. $\sec x$
• B. $3\sec x$
• C. $-3\sec x$
• D. $-\sec x$

Answer 1

The correct option is B $3\sec x$

Given,

$\frac{sinx}{siny}=\frac{1}{2}-\left(1\right)$

$\frac{cosx}{cosy}=\frac{3}{2}-\left(2\right),$

$x,y\in (0,\frac{\pi }{2})$

Divide eqs.(1) by (2)-

$\frac{tanx}{tany}=\frac{1}{3}$

$⟹\frac{\sqrt{1+se{c}^{2}x}}{\sqrt{1+se{c}^{2}y}}=\frac{1}{3}$

Squarring on both sides

$⟹\frac{1+se{c}^{2}x}{1+se{c}^{2}y}=\frac{1}{9}$

$⟹9+9se{c}^{2}x=1+se{c}^{2}y⟹8-se{c}^{2}y=-9se{c}^{2}x$

Taking square root on both sides

$\sqrt{se{c}^{2}y-8}=3secx$