If $8 sin θ = 4 + cos θ$ $8sintheta=4+costheta$ , what is the value of $tan θ$ $tantheta$ ?

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Answer 1 · 2018-07-14T08:43:27

$\to tan x = \frac{3}{4}$ $rarrtanx= 3/4$

Let $θ = x$ $theta=x$ then

$\to 8 sin x = 4 + cos x$ $rarr8sinx=4+cosx$

$\to \frac{8 sin x}{cos x} = \frac{4 + cos x}{cos x} = \frac{4}{cos x} + 1$ $rarr(8sinx)/cosx=(4+cosx)/cosx=4/cosx+1$

$\to 8 tan x = 4 sec x + 1$ $rarr8tanx=4secx+1$

$\to 8 tan x - 1 = 4 sec x$ $rarr8tanx-1=4secx$

Squaring both sides, we get,

$\to 64 {tan}^{2} x - 16 tan x + 1 = 16 {sec}^{2} x = 16 (1 + {tan}^{2} x)$ $rarr64tan^2x-16tanx+1=16sec^2x=16(1+tan^2x)$

$\to 64 {tan}^{2} x - 16 tan x + 1 = 16 + 16 {tan}^{2} x$ $rarr64tan^2x-16tanx+1=16+16tan^2x$

$\to 48 {tan}^{2} x - 16 tan x - 15 = 0$ $rarr48tan^2x-16tanx-15=0$

$\to 48 {tan}^{2} x + 20 tan x - 36 tan x - 15 = 0$ $rarr48tan^2x+20tanx-36tanx-15=0$

$\to 4 tan x (12 tan x + 5) - 3 (12 tan x + 5) = 0$ $rarr4tanx(12tanx+5)-3(12tanx+5)=0$

$\to (12 tan x + 5) (4 tan x - 3) = 0$ $rarr(12tanx+5)(4tanx-3)=0$

$\to tan x = - \frac{5}{12} or \frac{3}{4}$ $rarrtanx=-5/12 or 3/4$

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

If 8sintheta=4+costheta8sinθ=4+cosθ8sintheta=4+costheta, what is the value of tanthetatanθtantheta ?