Question

Which of the following shows the simplified form of sine of x over the quantity 1 minus cosine of x end quantity?

Answer 1

Solution

- The solution steps are given below:

$\begin{gathered} \frac{\sin x}{1-\cos x} \\ \\ \text{ Multiply both numerator and denominator by }(1+\cos x) \\ \\ =\frac{\sin x}{1-\cos x}\times\frac{(1+\cos x)}{(1+\cos x)} \\ \\ =\frac{\sin x+\cos x\sin x}{1-\cos^2x} \\ \\ \text{ But we have:} \\ \sin^2x=1-\cos^2x\text{ \lparen From Trigonometric identities\rparen} \\ \\ =\frac{\sin x+\cos x\sin x}{\sin^2x} \\ \\ =\frac{\sin x}{\sin^2x}+\frac{\cos x\sin x}{\sin^2x} \\ \\ =\frac{1}{\sin x}+\frac{\cos x}{\sin x} \\ \\ =\csc x+\cot x \end{gathered}$

Final Answer

The answer is:

$\csc x+\cot x$