Answer
See the full explanation below.
Work Step by Step
$\frac{\sin x}{\cos x+1}+\frac{\cos x-1}{\sin x}$
Multiply the numerator and denominator of $\frac{\sin x}{\cos x+1}$ by $\cos x-1$.
$\frac{\sin x}{\cos x+1}+\frac{\cos x-1}{\sin x}=\frac{\sin x}{\cos x+1}.\frac{\cos x-1}{\cos x-1}+\frac{\cos x-1}{\sin x}$
Multiply the above expression by using the formulae $(A+B)(A-B)={{A}^{2}}-{{B}^{2}}$ , with $A=1$ and $B=\cos x$ for the first numeric expression.
$\frac{\sin x}{\cos x+1}.\frac{\cos x-1}{\cos x-1}+\frac{\cos x-1}{\sin x}=\frac{\sin x\left( \cos x-1 \right)}{{{\cos }^{2}}x-1}+\frac{\cos x-1}{\sin x}$
Now, apply the Pythagorean identity of trigonometry: ${{\cos }^{2}}x-1=-{{\sin }^{2}}x$ , which comes out by solving ${{\sin }^{2}}x+{{\cos }^{2}}x=1$. Then, the above expression can be further simplified as:
$\frac{\sin x\left( \cos x-1 \right)}{{{\cos }^{2}}x-1}+\frac{\cos x-1}{\sin x}=\frac{\sin x\left( \cos x-1 \right)}{-{{\sin }^{2}}x}+\frac{\cos x-1}{\sin x}$
Multiply the above equation by $-$ to remove the negative sign from the denominator $-{{\sin }^{2}}x$.
$\begin{align}
& \frac{\sin x\left( \cos x-1 \right)}{-{{\sin }^{2}}x}+\frac{\cos x-1}{\sin x}=\frac{\sin x\left( \cos x-1 \right)}{-{{\sin }^{2}}x}+\frac{\cos x-1}{\sin x} \\
& =\frac{\sin x\left( \cos x-1 \right)}{-{{\sin }^{2}}x}+\frac{\cos x-1}{\sin x} \\
& =\frac{\sin x\left( 1-\cos x \right)}{{{\sin }^{2}}x}+\frac{\cos x-1}{\sin x} \\
& =\frac{1-\cos x}{\sin x}+\frac{\cos x-1}{\sin x}
\end{align}$
Take the least denominator as $sinx$ and add the constant terms in the numerator of the given expression as:
$\begin{align}
& \frac{1-\cos x}{\sin x}+\frac{\cos x-1}{\sin x}=\frac{1-\cos x+\cos x-1}{\sin x} \\
& =\frac{0}{\sin x} \\
& =0
\end{align}$
Thus, the left side of the expression is equal to the right side, which is
$\frac{\sin x}{\cos x+1}+\frac{\cos x-1}{\sin x}=0$.
