Question

if sin $\theta =\frac{4}{5}$ , Find the value of $\frac{4\tan \theta -5\cos \theta }{\sec \theta +4\cot \theta }$

Answer 1

We have
$sin\theta =\frac{4}{5}\Rightarrow \frac{perpendicular}{hypotenuse}=\frac{4}{5}$
so , we contract a right triangle ABC , right angle at B such that $\angle BAC=\theta$ , perpendicular = BC = 4 and Hypotenuse = AC = 5
By pythagoras theorem , we have
$A{C}^2=A{B}^2+B{C}^2$
$\Rightarrow .5^2=A{B}^2+4^2$
$\Rightarrow A{B}^2$ = 25 - 16 = 9
$\Rightarrow AB=\sqrt{9}=\Rightarrow$ AB = 3
$\therefore cos\theta =\frac{AB}{AC}=\frac{3}{5},tan\theta =\frac{BC}{AB}=\frac{4}{3}$
$cot\theta =\frac{AB}{AC}=\frac{3}{4},sec\theta =\frac{BC}{AB}=\frac{5}{3}$
$\therefore \frac{4tan\theta -5cos\theta }{sec\theta +4cot\theta }=\frac{4\times \frac{4}{3}-5\times \frac{3}{5}}{\frac{5}{3}+4\times \frac{3}{4}}=\frac{\frac{16}{3}-3}{\frac{5}{3}+3}=\frac{\frac{16-9}{3}}{\frac{5+9}{3}}=\frac{7}{14}=\frac{1}{2}$