Chapter 5.2, Problem 61E

Solution

To choose: The correct option from the given options.

The correct option is (E).

Given information:

The given identity is $\frac{\sin x}{1-\cos x}=\frac{1+\cos x}{\sin x}$ .

Formula used:

The trigonometric identity is ${\sin }^{2}x=1-{\cos }^{2}x$ .

The algebraic identity $\left(a-b\right)\left(a+b\right)=a^2-b^2$ .

Explanation:

The most important method to simplify a fraction is rationalizing the denominator.

Multiply the numerator and denominator of the left side of the given equation and simplify using algebraic and trigonometric identities.

  $\begin{array}{c}\frac{\sin x}{1-\cos x}=\frac{\sin x}{\left(1-\cos x\right)}\cdot \frac{\left(1+\cos x\right)}{\left(1+\cos x\right)}\\ =\frac{\sin x\left(1+\cos x\right)}{1-{\cos }^{2}x}\\ =\frac{\sin x\left(1+\cos x\right)}{{\sin }^{2}x}\\ =\frac{1+\cos x}{\sin x}\end{array}$

This implies that the efficient first to prove the identity is $\frac{\sin x}{1-\cos x}=\frac{\sin x}{\left(1-\cos x\right)}\cdot \frac{\left(1+\cos x\right)}{\left(1+\cos x\right)}$ .

Therefore, the correct option is (E).