Question

If $\theta$ is the measure of an angle such that $b\sin \theta =a\cos \theta$, then $\frac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta }$ is..

• A. $\frac{a^2+b^2}{a^2-b^2}$
• B. $\frac{a^2-b^2}{a^2+b^2}$
• C. $\frac{a+b}{a-b}$
• D. $\frac{a-b}{a+b}$

Answer 1

The correct option is B $\frac{a^2-b^2}{a^2+b^2}$

$b\sin \theta =a\cos \theta$

Squaring on both sides

$b^2{\sin }^2\theta =a^2{\cos }^2\theta$

$⟹b^2{\sin }^2\theta =a^2(1-{\sin }^2\theta )$

$⟹{\sin }^2\theta =\frac{a^2}{a^2+b^2}$

$⟹\sin \theta =\frac{a}{\sqrt{a^2+b^2}}$

and

$\cos \theta =\frac{b}{\sqrt{a^2+b^2}}$

$\frac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta }=\frac{a\left(\frac{a}{{\sqrt{a}}^2+b^2}\right)-b\left(\frac{b}{\sqrt{a^2+b^2}}\right)}{a\frac{a}{\sqrt{a^2+b^2}}+b\left(\frac{b}{\sqrt{}{b}^2+a^2}\right)}=\frac{a^2-b^2}{a^2+b^2}$