Question

If $5\tan \theta =3$, then $\frac{5\tan \theta -3\cos \theta }{5\tan \theta +3\cos \theta }=$ ?

• A. $0$
• B. $\frac{5}{3}$
• C. $\frac{3}{5}$
• D. $\frac{4}{5}$
• E. No solution

Answer 1

Answer: (E)

No solution

$5\tan \theta =3⟹\tan \theta =\frac{3}{5}⟹P=3,B=5$

$\because H=\sqrt{P^2+B^2}\therefore H=\sqrt{9+25}=\sqrt{34}$

$\therefore \cos \theta =\frac{B}{H}=\frac{5}{\sqrt{34}}⟹3\cos \theta =\frac{15}{\sqrt{34}}$

$\therefore \frac{5\tan \theta -3\cos \theta }{5\tan \theta +3\cos \theta }=\frac{3-\frac{15}{\sqrt{34}}}{3+\frac{15}{\sqrt{34}}}=\frac{\sqrt{34}-5}{\sqrt{34}+5}$

Hence Answer choice E (No Solution) is correct