Question #545b9

Oct 21, 2016

$\cos \left(x\right) \left(\tan \left(x\right) + \cot \left(x\right)\right) = \frac{1}{\sin} \left(x\right)$

Explanation:

We will use the definitions of the tangent and cotangent functions:

• $\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right)$
• $\cot \left(x\right) = \cos \frac{x}{\sin} \left(x\right)$

as well as the identity ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$
Note that subtracting ${\sin}^{2} \left(x\right)$ from both sides produces the form we will use:

• ${\cos}^{2} \left(x\right) = 1 - {\sin}^{2} \left(x\right)$

With those, we have

$\cos \left(x\right) \left(\tan \left(x\right) + \cot \left(x\right)\right) = \cos \left(x\right) \left(\sin \frac{x}{\cos} \left(x\right) + \cos \frac{x}{\sin} \left(x\right)\right)$

$= \cancel{\cos \left(x\right)} \cdot \sin \frac{x}{\cancel{\cos \left(x\right)}} + \cos \left(x\right) \cdot \cos \frac{x}{\sin} \left(x\right)$

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

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

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

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

$= \frac{1}{\sin} \left(x\right)$