Question

If $3\cos \theta =5\sin \theta$, then the value of $\frac{5\sin \theta -2{\sec }^3\theta +2\cos \theta }{5\sin \theta -2{\sec }^3\theta +2\cos \theta }$ is ?

Answer 1

$3\cos \theta =5\sin \theta$

$\cos \theta =\left(\frac{5}{3}\right)\sin \theta$

${\cos }^2\theta ={\left(\frac{5}{3}\right)}^2{\sin }^2\theta$

$1-{\sin }^2\theta =\left(\frac{25}{9}\right){\sin }^2\theta$

$\Rightarrow {\sin }^2\theta =\frac{9}{34}$

$\therefore \sin \theta =\frac{3}{\sqrt{34}}$

${\cos }^2\theta =1-{\sin }^2\theta =1-\frac{9}{34}=\frac{25}{34}$

$\therefore \cos \theta =\frac{5}{\sqrt{34}}$

$\therefore \frac{5\sin \theta -2{\sec }^3\theta +2\cos \theta }{5\sin \theta -2{\sec }^3\theta +2\cos \theta }$

$=\frac{5\times \frac{3}{\sqrt{34}}-2\times \frac{34\sqrt{34}}{125}+2\times \frac{5}{\sqrt{34}}}{5\times \frac{3}{\sqrt{34}}-2\times \frac{34\sqrt{34}}{125}+2\times \frac{5}{\sqrt{34}}}$

$=\frac{\frac{15}{\sqrt{34}}-2\times \frac{34\sqrt{34}}{125}+\frac{10}{\sqrt{34}}}{\frac{15}{\sqrt{34}}-2\times \frac{34\sqrt{34}}{125}+\frac{10}{\sqrt{34}}}$

$=\frac{\frac{25}{\sqrt{34}}-2\times \frac{34\sqrt{34}}{125}}{\frac{25}{\sqrt{34}}-2\times \frac{34\sqrt{34}}{125}}$

$=1$