Question

If tan $\theta =\frac{a}{b},$ prove that $\frac{asin\theta -bcos\theta }{asin\theta +bcos\theta }=\frac{a^2-b^2}{a^2+b^2}$

Answer 1

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

$\frac{\sin \theta }{\cos \theta }=\frac{a}{b}$

$\sin \theta =\frac{a\cos \theta }{b}$

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

Hence Proved.