If 3 cos θ = 5 sin θ, then the value of 5 sin θ−2 sec3 θ+2 cos θ5 sin θ+2 sec3 θ−2 cos θis
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

(a) $\frac{271}{979}$

(b) $\frac{316}{2937}$

(c) $\frac{542}{2937}$

(d) None of these

Solution:

We have,

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

So we can manipulate it as,

$\tan \theta=\frac{3}{5}$

So now we can get the values of other trigonometric ratios,

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

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

$\sec \theta=\frac{\sqrt{34}}{5}$

So now we will put these values in the equation,

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

$=\frac{5\left(\frac{3}{\sqrt{34}}\right)-2\left(\frac{34 \sqrt{34}}{125}\right)+\frac{10}{\sqrt{34}}}{5\left(\frac{3}{\sqrt{34}}\right)+2\left(\frac{34 \sqrt{34}}{125}\right)-\frac{10}{\sqrt{34}}}$

$=\frac{(15)(125)-(2)(34)^{2}+1250}{(15)(125)+(2)(34)^{2}-1250}$

$=\frac{271}{979}$