# How do you prove sin^2x/ (cos^2 +3cosx+2) = (1-cosx)/ (2+cosx)?

Apr 16, 2018

As proved.

#### Explanation:

${\sin}^{2} \frac{x}{{\cos}^{2} x + 3 \cos x + 2}$

$\implies \frac{1 - {\cos}^{2} x}{{\cos}^{2} x + \cos x + 2 \cos x + 2}$

$\implies \frac{\left(1 + \cos x\right) \cdot \left(1 - \cos x\right)}{\cos x \left(\cos x + 1\right) + 2 \cdot \left(\cos x + 1\right)}$

$\implies \frac{\left(\cancel{1 + \cos x}\right) \cdot \left(1 - \cos x\right)}{\left(\cancel{\cos x + 1}\right) \left(\cos x + 2\right)}$

$\implies \frac{1 - \cos x}{2 + \cos x} = R H S$

Q E D

Apr 16, 2018

Kindly refer to a Proof in the Explanation.

#### Explanation:

${\sin}^{2} \frac{x}{{\cos}^{2} x + 3 \cos x + 2}$,

$= \frac{1 - {\cos}^{2} x}{\left(\cos x + 1\right) \left(\cos x + 2\right)}$,

$= \frac{\cancel{\left(1 + \cos x\right)} \left(1 - \cos x\right)}{\cancel{\left(\cos x + 1\right)} \left(\cos x + 2\right)}$,

$= \frac{1 - \cos x}{2 + \cos x}$, as desired!

Apr 16, 2018

$\text{see explanation}$

#### Explanation:

$\text{using the "color(blue)"trigonometric identity}$

â€¢color(white)(x)sin^2x+cos^2x=1

$\Rightarrow {\sin}^{2} x = 1 - {\cos}^{2} x$

$\text{consider the left side}$

$\frac{{\sin}^{2} x}{{\cos}^{2} x + 3 \cos x + 2}$

$= \frac{1 - {\cos}^{2} x}{{\cos}^{2} x + 3 \cos x + 2}$

$\text{the numerator is a "color(blue)"difference of squares}$

$\text{and the denominator is a quadratic in cos}$

$= \frac{\left(1 - \cos x\right) \cancel{\left(1 + \cos x\right)}}{\left(\cos x + 2\right) \cancel{\left(\cos x + 1\right)}}$

$= \frac{1 - \cos x}{2 + \cos x} = \text{ right side "rArr"proven}$

Apr 16, 2018

After replacing :${\cos}^{2.} \to {\cos}^{2} x$

#### Explanation:

Here,

${\sin}^{2} \frac{x}{{\cos}^{2} \textcolor{red}{x} + 3 \cos x + 2} = \frac{1 - \cos x}{2 + \cos x}$

Now,

${\cos}^{2} x + 3 \cos x + 2 = {\cos}^{2} x + 2 \cos x + \cos x + 2$

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} = \cos x \left(\cos x + 2\right) + 1 \left(\cos x + 2\right)$

$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots} = \left(\cos x + 1\right) \left(\cos x + 2\right)$

So,

$L H S = {\sin}^{2} \frac{x}{{\cos}^{2} x + 3 \cos x + 2}$

$= \frac{1 - {\cos}^{2} x}{\left(\cos x + 1\right) \left(\cos x + 2\right)}$

$= \frac{\left(1 - \cos x\right) \cancel{\left(1 + \cos x\right)}}{\cancel{\left(1 + \cos x\right)} \left(2 + \cos x\right)}$

$= \frac{1 - \cos x}{2 + \cos x}$

$= R H S$