Question

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

• A. $\frac{\sqrt{a^2+b^2-c^2}}{c}$
• B. $-\frac{\sqrt{a^2+b^2-c^2}}{c}$
• C. $\pm \frac{\sqrt{a^2+b^2-c^2}}{c}$
• D. $\pm \frac{\sqrt{a^2+b^2+c^2}}{c}$

Answer 1

Answer: (C)

$\pm \frac{\sqrt{a^2+b^2-c^2}}{c}$

$a\sin \theta +b\cos \theta =c$

$a\cos \theta -b\sin \theta =k$

Squaring on both sides and adding gives

$a^2+b^2=c^2+k^2$

$k=\pm \sqrt{a^2+b^2-c^2}$

$\frac{a-b\tan \theta }{b+a\tan \theta }=\frac{a\cos \theta -b\sin \theta }{a\sin \theta +b\cos \theta }=\frac{k}{c}=\frac{\pm \sqrt{a^2+b^2-c^2}}{c}$