Question

If$5tan\Theta -4=0,$ then find the value of the following expression : $\frac{5sin\Theta -3cos\Theta }{5sin\Theta +2cos\Theta }.$

Answer 1

$5\tan \theta =4$

\tan\theta=\cfrac{4}{5}$

$⟹\sin \theta =\frac{4}{\sqrt{25+16}}=\frac{4}{\sqrt{41}}$

$⟹\cos \theta =\frac{5}{\sqrt{25+26}}=\frac{5}{\sqrt{41}}$

$\frac{5\sin \theta -3\cos \theta }{5\sin \theta +2\cos \theta }=\frac{5\left(4/\sqrt{41}\right)-3\left(5/\sqrt{41}\right)}{5\left(4/\sqrt{41}\right)+2\left(5/\sqrt{41}\right)}=\frac{5}{30}=\frac{1}{6}$