Question

Which of the following shows the simplified form of (sin x)/(1 - cos x)? 1sin x + tan xsin x + cot xcsc x + cot x

Answer 1

We want to simplify the following expression

$\frac{\sin x}{1-\cos x}$

We can start by multiplying both numerator and denominator by the conjugate of the denominator:

$\begin{gathered} \frac{\sin(x)}{1-\cos(x)}=\frac{\sin(x)}{1-\cos(x)}\cdot\frac{1+\cos(x)}{1+\cos(x)} \\ \\ =\frac{\sin(x)(1+\cos(x))}{(1-\cos(x))(1+\cos(x))} \\ \\ =\frac{\sin(x)\cdot(1)+\sin(x)\cdot\cos(x)}{(1)\cdot(1)+(1)\cdot(\cos(x))+(-\cos(x))\cdot(1)+(-\cos(x))\cdot(\cos(x))} \\ \\ \begin{equation*} =\frac{\sin(x)+\sin(x)\cos(x)}{1-\cos^2(x)} \end{equation*} \end{gathered}$

Then, using the identity

$\sin^2x+\cos^2x=1\implies\sin^2x=1-\cos^2x$

We can rewrite our expression as

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

By definition of cosecant and cotangent, our expression can be written as

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

and this is our answer.

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