Final answer to the problem
Step-by-step Solution
How should I solve this problem?
- Prove from RHS (right-hand side)
- Prove from LHS (left-hand side)
- Express everything into Sine and Cosine
- Exact Differential Equation
- Linear Differential Equation
- Separable Differential Equation
- Homogeneous Differential Equation
- Integrate by partial fractions
- Product of Binomials with Common Term
- FOIL Method
- Load more...
Starting from the right-hand side (RHS) of the identity
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 $\sin\left(x\right)$
Since we have reached the expression of our goal, we have proven the identity