Question

If $\cot \theta =\frac{b}{a},$ then $\frac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta }=$

Answer 1

$\cot \theta =\frac{b}{a}$

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

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