Question

If $\cos \alpha =\frac{1}{2}(x+\frac{1}{x}),\cos \beta =\frac{1}{2}(y+\frac{1}{y})$ then $\cos (\alpha -\beta )$ is equal to

• A. $\frac{x}{y}+\frac{y}{x}$
• B. $xy+\frac{1}{xy}$
• C. $\frac{1}{2}(\frac{x}{y}+\frac{y}{x})$
• D. none of these

Answer 1

The correct option is C $\frac{1}{2}(\frac{x}{y}+\frac{y}{x})$

$\cos \alpha =\frac{1}{2}(x+\frac{1}{x})=\frac{x^2+1}{2x}$

$\Rightarrow \sin \alpha =\frac{(x^2-1)}{2x}$

$\cos \beta =\frac{1}{2}(y+\frac{1}{y})=\frac{y^2+1}{2y}$

$\Rightarrow \sin \beta =\frac{(y^2-1)}{2y}$

$\cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta$

$=\frac{x^2+1}{2x}\frac{y^2+1}{2y}+\frac{(x^2-1)}{2x}\frac{(y^2-1)}{2y}$

$=\frac{x^2+y^2}{2xy}=\frac{1}{2}(\frac{x}{y}+\frac{y}{x})$