Question

If $\sec \theta =\frac{13}{5},\text{then show that}\frac{2sin\theta -3cos\theta }{4sin\theta -9cos\theta }=3$

Answer 1

$\sec \theta =\frac{13}{5}=\frac{hypotenuse}{adjacentside}$
Hypotenuse= 13; Adjacent side = 5
By pythogoras theorem,
$\text{Opposite side}=\sqrt{(hypotenuse{)}^2-(adjacentside{)}^2}=\sqrt{(13{)}^2-(5{)}^2}=12\sin \theta =\frac{oppositeside}{hypotenuse}=\frac{12}{13}\cos \theta =\frac{adjacentside}{hypotenuse}=\frac{5}{13}Now,\frac{2sin\theta -3cod\theta }{4sin\theta -9cos\theta }=\frac{2\left(\frac{12}{13}\right)-3\left(\frac{5}{13}\right)}{4\left(\frac{12}{13}\right)-9\left(\frac{5}{13}\right)}=\frac{\frac{24}{13}-\frac{15}{13}}{\frac{48}{13}-\frac{45}{13}}=\frac{\frac{9}{13}}{\frac{3}{13}}=\frac{9}{3}=3$
Hence proved.