Question

If θ = $\frac{a}{b}$, then $\frac{(a\mathrm{sin\theta }-b\mathrm{cos\theta })}{(a\sin \theta +b\mathrm{cos\theta })}$ = ?
(a) $\frac{(a^2+b^2)}{(a^2-b^2)}$
(b) $\frac{(a^2-b^2)}{(a^2+b^2)}$
(c) $\frac{a^2}{(a^2+b^2)}$
(d) $\frac{b^2}{(a^2+b^2)}$

Answer 1

(b) $\frac{\left(a^2-b^2\right)}{\left(a^2+b^2\right)}$
We have tan θ = $\frac{a}{b}$

Now, dividing the numerator and denominator of the given expression by cos θ, we get:
$$\begin{gathered} \frac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta } \\ =\frac{{\displaystyle \frac{1}{\cos \theta }}\left(a\sin \theta -b\cos \theta \right)}{{\displaystyle \frac{1}{\cos \theta }}\left(a\sin \theta +b\cos \theta \right)}=\frac{a\tan \theta -b}{a\tan \theta +b}=\frac{{\displaystyle \frac{a^2}{b}}-b}{{\displaystyle \frac{a^2}{b}}+b}=\frac{a^2-b^2}{a^2+b^2} \end{gathered}$$