Question

If $\cos {\theta }_1=\frac{1}{2}(x+\frac{1}{x})$ , $\cos {\theta }_2=\frac{1}{2}(y+\frac{1}{y})$ then
$\cos ({\theta }_1-{\theta }_2)$ equals

• A. $xy+\frac{1}{xy}$
• B. $\frac{x}{y}+\frac{y}{x}$
• C. $\frac{1}{2}\left(\frac{x^2+y^2}{xy}\right)$
• D. $\frac{1}{2}\left(\frac{x^2-y^2}{xy}\right)$

Answer 1

Answer: (C)

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

$\cos {\theta }_1=\frac{1}{2}(x+\frac{1}{x})$

$\Rightarrow x=e^{i{\theta }_1}$
And $\cos {\theta }_2=\frac{1}{2}(y+\frac{1}{y})$

$\Rightarrow y=e^{i{\theta }_2}$
$\Rightarrow \frac{x}{y}=e^{i\left({\theta }_1{-\theta }_2\right)}$ ...(1)

And $\frac{y}{x}=e^{-i\left({\theta }_1{-\theta }_2\right)}$ ...(2)
Adding (1) and (2), we get
$\frac{x}{y}+\frac{y}{x}=e^{i\left({\theta }_1{-\theta }_2\right)}+e^{-i\left({\theta }_1{-\theta }_2\right)}$
$\Rightarrow 2\cos \left({\theta }_1{-\theta }_2\right)=\frac{x}{y}+\frac{y}{x}\therefore \cos \left({\theta }_1{-\theta }_2\right)=\frac{x^2{+y}^2}{2xy}$
Ans: C