Question

If $\tan \theta =\frac{x}{y},$ then $\frac{x\sin \theta +y\cos \theta }{x\sin \theta -y\cos \theta }$ is equal to

• A. $\frac{x^2+y^2}{x^2-y^2}$
• B. $\frac{x^2-y^2}{x^2+y^2}$
• C. $\frac{x}{\sqrt{x^2+y^2}}$
• D. $\frac{y}{\sqrt{x^2+y^2}}$

Answer 1

The correct option is A $\frac{x^2+y^2}{x^2-y^2}$

$\frac{x\sin \theta +y\cos \theta }{x\sin \theta -y\cos \theta }$

Multiply and divide by $\cos \theta$
$=\frac{x\frac{\sin \theta }{\cos \theta }+y\frac{\cos \theta }{\cos \theta }}{x\frac{\sin \theta }{\cos \theta }-y\frac{\cos \theta }{\cos \theta }}$

$=\frac{x\times \frac{x}{y}+y}{x\times \frac{x}{y}-y}$
$=\frac{\frac{x^2+y^2}{y}}{\frac{x^2-y^2}{y}}$
$=\frac{x^2+y^2}{x^2-y^2}$