Question

If $\tan \theta =\frac{a}{b}$, then $\frac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos \theta }=.....$

• 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

The correct option is B $\frac{a^2-b^2}{a^2+b^2}$

Dividing numerator and denominator by $\cos \theta$ we get
$\frac{(a\sin \theta -b\cos \theta )}{(a\sin \theta +b\cos \theta )}=\frac{a\tan \theta -b}{a\tan \theta +b}=\frac{a\times \frac{a}{b}-b}{a\times \frac{a}{b}+b}=\frac{a^2-b^2}{a^2+b^2}$