Question

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

Answer 1

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

To show: $\frac{asin\theta -bcos\theta }{asin\theta +bcos\theta }=\frac{a^2-b^2}{a^2+b^2}$

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

$\Rightarrow \frac{sin\theta }{cos\theta }=\frac{a}{b}\Rightarrow bsin\theta =acos\theta =\lambda \left(say\right)\Rightarrow sin\theta =\frac{\lambda }{b}andcos\theta =\frac{\lambda }{a}$

How, $\frac{asin\theta -bcos\theta }{asin\theta +bcos\theta }=\frac{\frac{a.\lambda }{b}-\frac{b.\lambda }{a}}{\frac{a.\lambda }{b}+\frac{b.\lambda }{a}}$

$=\frac{\lambda (\frac{a}{b}-\frac{b}{a})}{\lambda (\frac{a}{b}+\frac{b}{a})}=\frac{\frac{a}{b}-\frac{b}{a}}{\frac{a}{b}+\frac{b}{a}}=\frac{\frac{a^2-b^2}{ab}}{\frac{a^2+b^2}{ab}}=\frac{a^2-b^2}{a^2+b^2}$

Hence proved.