# How do you verify sin x + cos x * cot x = csc x?

##### 1 Answer
Mar 2, 2016

Recall the following reciprocal, tangent, and Pythagorean identities:

$1$. $\textcolor{\mathmr{and} a n \ge}{\cot} x = \frac{1}{\tan} x$

$2$. $\textcolor{b l u e}{\csc} x = \frac{1}{\sin} x$

$3$. $\textcolor{p u r p \le}{\tan} x = \sin \frac{x}{\cos} x$

$4$. $\textcolor{b r o w n}{{\sin}^{2} x + {\cos}^{2} x} = 1$

Proving the Identity
$1$. Simplify by rewriting the left side of the identity in terms of $\sin x$ and $\cos x$.

$\sin x + \cos x \cdot \textcolor{\mathmr{and} a n \ge}{\cot} x = \textcolor{b l u e}{\csc} x$

Left side:

$\sin x + \cos x \cdot \frac{1}{\textcolor{p u r p \le}{\tan}} x$

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

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

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

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

$2$. Simplify by multiplying $\cos x$ by $\cos \frac{x}{\sin} x$.

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

$3$. Find the L.C.M. (lowest common multiple) to rewrite the left hand side as a fraction.

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

$4$. Simplify.

$= \frac{\textcolor{b r o w n}{\left({\sin}^{2} x + {\cos}^{2} x\right)}}{\sin} x$

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

$= \textcolor{g r e e n}{\csc x}$

$\therefore$, left side$=$right side.