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. $\frac{ab}{a^2+b^2}$
• C. $\frac{a^2{b}^2}{a+b}$
• D. None of these

Answer 1

Answer: (D)

None of these

$\tan \theta =\frac{a}{b},$ then $\frac{a\sin \theta -b\cos \theta }{a\sin \theta +b\cos }=\frac{\frac{a\sin \theta }{b\cos \theta }-1}{\frac{a\sin \theta }{b\cos \theta }+1}$

$=\frac{\frac{a}{b}\tan \theta -1}{\frac{a}{b}\tan \theta +1}$

$=\frac{\frac{a^2}{b^2}-1}{\frac{a^2}{b^2}+1}$

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