Solve this


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



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 θ">θθ in the above expression, we get:




 i.e., LHS = RHS

 Hence proved.

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