# Question #5d572

Dec 26, 2016

Using the definitions of $\csc$ and $\cot$, along with the identities

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

we have

$\csc \left(2 x\right) + \cot \left(2 x\right) = \frac{1}{\sin} \left(2 x\right) + \cos \frac{2 x}{\sin} \left(2 x\right)$

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

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

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

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

$= \cot \left(x\right)$

Dec 26, 2016

See proof below

#### Explanation:

We use

$\csc x = \frac{1}{\sin} x$

$\sin 2 x = 2 \sin x \cos x$

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

$\cot x = \cos \frac{x}{\sin} x$

So,

$\csc 2 x + \cot 2 x = \frac{1}{\sin 2 x} + \frac{\cos 2 x}{\sin 2 x}$

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

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

$= \frac{2 {\cos}^{2} x}{2 \sin x \cos x}$

$= \cos \frac{x}{\sin} x$

$= \cot x$

Q.E.D