Question

If tan θ = $\frac{a}{b}$, show that $\frac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta }=\frac{a^2-b^2}{a^2+b^2}$.

Answer 1

$$\begin{gathered} \text{Given: tan}\theta \text{=}\frac{a}{b} \\ \text{LHS=}\frac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta } \\ \text{=}\frac{a\tan \theta -b}{a\tan \theta +b}\left(\begin{array}{l}\text{Dividing both numerator and}\\ \text{denominator by cos}\theta \end{array}\right) \\ \text{=}\frac{a\times \frac{a}{b}-b}{a\times \frac{a}{b}+b} \\ \text{=}\frac{\frac{a^2}{b}-b}{\frac{a^2}{b}+b} \\ \text{=}\frac{a^2-b^2}{a^2+b^2} \end{gathered}$$

Hence, LHS = RHS