Final answer to the problem
Step-by-step Solution
Specify the solving method
I. Express the LHS in terms of sine and cosine and simplify
Start from the LHS (left-hand side)
Multiply and divide the fraction $\frac{1+\cos\left(x\right)}{\sin\left(x\right)}$ by the conjugate of it's numerator $1+\cos\left(x\right)$
Multiplying fractions $\frac{1+\cos\left(x\right)}{\sin\left(x\right)} \times \frac{1-\cos\left(x\right)}{1-\cos\left(x\right)}$
The sum of two terms multiplied by their difference is equal to the square of the first term minus the square of the second term. In other words: $(a+b)(a-b)=a^2-b^2$.
Apply the trigonometric identity: $1-\cos\left(\theta \right)^2$$=\sin\left(\theta \right)^2$
Simplify the fraction $\frac{\sin\left(x\right)^2}{\sin\left(x\right)\left(1-\cos\left(x\right)\right)}$ by $\sin\left(x\right)$
II. Express the RHS in terms of sine and cosine and simplify
Start from the RHS (right-hand side)
Multiply and divide the fraction $\frac{\sin\left(x\right)}{1-\cos\left(x\right)}$ by the conjugate of it's denominator $1-\cos\left(x\right)$
Multiplying fractions $\frac{\sin\left(x\right)}{1-\cos\left(x\right)} \times \frac{1+\cos\left(x\right)}{1+\cos\left(x\right)}$
The sum of two terms multiplied by their difference is equal to the square of the first term minus the square of the second term. In other words: $(a+b)(a-b)=a^2-b^2$.
Apply the trigonometric identity: $1-\cos\left(\theta \right)^2$$=\sin\left(\theta \right)^2$
Simplify the fraction by $\sin\left(x\right)$
III. Choose what side of the identity are we going to work on
To prove an identity, we usually begin to work on the side of the equality that seems to be more complicated, or the side that is not expressed in terms of sine and cosine. In this problem, we will choose to work on the right side $\frac{1+\cos\left(x\right)}{\sin\left(x\right)}$ to reach the left side $\frac{\sin\left(x\right)}{1-\cos\left(x\right)}$
Multiply and divide the fraction $\frac{1+\cos\left(x\right)}{\sin\left(x\right)}$ by the conjugate of it's numerator $1+\cos\left(x\right)$
Multiplying fractions $\frac{1+\cos\left(x\right)}{\sin\left(x\right)} \times \frac{1-\cos\left(x\right)}{1-\cos\left(x\right)}$
The sum of two terms multiplied by their difference is equal to the square of the first term minus the square of the second term. In other words: $(a+b)(a-b)=a^2-b^2$.
Apply the trigonometric identity: $1-\cos\left(\theta \right)^2$$=\sin\left(\theta \right)^2$
Simplify the fraction $\frac{\sin\left(x\right)^2}{\sin\left(x\right)\left(1-\cos\left(x\right)\right)}$ by $\sin\left(x\right)$
IV. Check if we arrived at the expression we wanted to prove
Since we have reached the expression of our goal, we have proven the identity