Question: If $tan \theta$ = 2, find the value of:
$\dfrac{8sin\theta + 5cos\theta}{sin^3\theta + 2cos^3\theta + 3cos\theta}$


Answer: 21/5

Solution:
Given,
$tan \theta$ = 2
We know,
$tan \theta = \dfrac{sin \theta}{cos \theta}$
$or, 2 = \dfrac{sin \theta}{cos \theta}$
$or, sin \theta = 2 cos \theta$ - (i)

Also,
$ sin 2\theta = \dfrac{2 tan \theta }{1 + tan^2 \theta}$

[Put value of tan $\theta$]
$or, 2 sin \theta cos \theta = \dfrac{2 × 2 }{1 + 2^2}$

[Put value of sin$\theta$ from equation (i)]
$or, 2 × (2cos \theta) cos \theta = \dfrac{4}{1 + 4}$

$or, 4 cos^2 \theta = \dfrac{4}{5}$

$or, cos^2 \theta = \dfrac{1}{5}$

Given,
$\dfrac{8sin\theta + 5cos\theta}{sin^3\theta + 2cos^3\theta + 3cos\theta}$

$= \dfrac{8(2cos\theta) + 5cos \theta}{(2cos\theta)^3 + 2cos^3\theta + 3cos\theta}$

$= \dfrac{16 cos\theta + 5cos\theta}{8cos^3\theta + 2cos^3 \theta + 3cos\theta}$

$= \dfrac{21 cos\theta}{10cos^3 \theta + 3cos\theta)}$

$= \dfrac{21}{10 cos^3\theta + 3}$

$= \dfrac{21 cos\theta}{cos\theta(10cos^2\theta + 3}$

$= \dfrac{21}{10(1/5) + 3}$

$= \dfrac{21}{2+3}$

$= \dfrac{21}{5}$