If 8sintheta=4+costheta, what is the value of tantheta ?

Jul 14, 2018

$\rightarrow \tan x = \frac{3}{4}$

Explanation:

Let $\theta = x$ then

$\rightarrow 8 \sin x = 4 + \cos x$

$\rightarrow \frac{8 \sin x}{\cos} x = \frac{4 + \cos x}{\cos} x = \frac{4}{\cos} x + 1$

$\rightarrow 8 \tan x = 4 \sec x + 1$

$\rightarrow 8 \tan x - 1 = 4 \sec x$

Squaring both sides, we get,

$\rightarrow 64 {\tan}^{2} x - 16 \tan x + 1 = 16 {\sec}^{2} x = 16 \left(1 + {\tan}^{2} x\right)$

$\rightarrow 64 {\tan}^{2} x - 16 \tan x + 1 = 16 + 16 {\tan}^{2} x$

$\rightarrow 48 {\tan}^{2} x - 16 \tan x - 15 = 0$

$\rightarrow 48 {\tan}^{2} x + 20 \tan x - 36 \tan x - 15 = 0$

$\rightarrow 4 \tan x \left(12 \tan x + 5\right) - 3 \left(12 \tan x + 5\right) = 0$

$\rightarrow \left(12 \tan x + 5\right) \left(4 \tan x - 3\right) = 0$

$\rightarrow \tan x = - \frac{5}{12} \mathmr{and} \frac{3}{4}$

But $x = {\tan}^{- 1} \left(- \frac{5}{12}\right)$ does not satisfy the equation so $\tan x = \frac{3}{4}$