Question

If $\tan \theta =\frac{x}{y}$, then the value of $\left(\frac{x\sin \theta +y\cos \theta }{x\sin \theta -y\cos \theta }\right)$

• 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

Answer: (A)

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

$=\left(\frac{x\sin \theta +y\cos \theta }{x\sin \theta -y\cos \theta }\right)$

Divide by $\cos \theta$
$=\left(\frac{x\tan \theta +y}{x\tan \theta -y}\right)$
$\tan \theta =\frac{x}{y}......given$
$=\frac{\frac{x^2}{y}+y}{\frac{x^2}{y}-y}$
$=\frac{x^2+y^2}{x^2-y^2}$