Solve this
Question:

If $\tan \theta=\frac{a}{b}$, show that $\left(\frac{a \sin \theta-b \cos \theta}{a \sin \theta+b \cos \theta}\right)=\frac{\left(a^{2}-b^{2}\right)}{\left(a^{2}+b^{2}\right)}$

Solution:

It is given that $\tan \theta=\frac{a}{b}$.

$\mathrm{LHS}=\frac{a \sin \theta-b \cos \theta}{a \sin \theta+b \cos \theta}$

Dividing the numerator and denominator by $\cos \theta$, we get:

$\frac{a \tan \theta-b}{a \tan \theta+b} \quad\left(\because \tan \theta=\frac{\sin \theta}{\cos \theta}\right)$

Now, substituting the value of tan θθ in the above expression, we get:

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

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

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

i.e., LHS = RHS

Hence proved.