Question

If tan θ = ${\displaystyle \frac{a}{b}}$ prove that ${\displaystyle \frac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta }=\frac{a^2-b^2}{a^2+b^2}}$

Answer 1

Let  ${\displaystyle \frac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta }}$

Divide both Nr and Dr with cos θ of (a)

${\displaystyle \frac{\frac{a\sin \theta -b\cos \theta }{\cos \theta }}{\frac{a\sin \theta +b\cos \theta }{\cos \theta }}}$

${\displaystyle =\frac{\tan \theta -b}{a\tan \theta +b}}$

${\displaystyle =\frac{a\times \left(\frac{a}{b}\right)-b}{a\times \left(\frac{a}{b}\right)+b}}$

${\displaystyle =\frac{a^2-b^2}{a^2+b^2}}$