How do you simplify the expression $\frac{1 + cos x}{sin x} + \frac{sin x}{1 + cos x}$ $(1+cosx)/sinx+sinx/(1+cosx)$ ?

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How do you simplify the expression $\frac{1 + cos x}{sin x} + \frac{sin x}{1 + cos x}$ $(1+cosx)/sinx+sinx/(1+cosx)$ ?

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Answer 1 · 2016-08-30T18:35:02

The expression can be simplified to $2 csc x$ $2cscx$

Explanation:

Start by putting on a common denominator.

$\Rightarrow \frac{(1 + cos x) (1 + cos x)}{(sin x) (1 + cos x)} + \frac{sin x (sin x)}{(sin x) (1 + cos x)}$ $=>((1 + cosx)(1 + cosx))/((sinx)(1 + cosx)) + (sinx(sinx))/((sinx)(1 + cosx))$

$\Rightarrow \frac{{cos}^{2} x + 2 cos x + 1 + {sin}^{2} x}{sin x (1 + cos x)}$ $=>(cos^2x + 2cosx + 1 + sin^2x)/(sinx(1 + cosx))$

Apply the pythagorean identity ${cos}^{2} x + {sin}^{2} x = 1$ $cos^2x + sin^2x = 1$ :

$\Rightarrow \frac{1 + 2 cos x + 1}{sin x (1 + cos x)}$ $=>(1 + 2cosx + 1)/(sinx(1 + cosx)$

$\Rightarrow \frac{2 + 2 cos x}{sin x (1 + cos x)}$ $=>(2 + 2cosx)/(sinx(1 + cosx))$

Factor out a $2$ $2$ in the numerator.

$\Rightarrow \frac{2 (1 + cos x)}{sin x (1 + cos x)}$ $=> (2(1 + cosx))/(sinx(1 + cosx))$

$\Rightarrow \frac{2}{sin x}$ $=>2/sinx$

Finally, apply the reciprocal identity $\frac{1}{sin θ} = csc θ$ $1/sintheta = csctheta$ to get:

$\Rightarrow 2 csc x$ $=> 2cscx$

Hopefully this helps!

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