Trigonometric Proofs Exercise 3

When trying to prove an identity with fractions, it’s often useful to combine the fractions using a common denominator. Like, in this case, everything is in terms of sine and cosine, so combining fractions is the best step forward.

The common denominator of the fractions on the LHS will be the product of the denominators, \((1+\cos{x})\sin{x}\). So we need to multiply both the fractions by the missing factor, \(\sin{x}\) and \((1+\cos{x})\) respectively, to achieve the common denominator. AND we have to multiply the numerators by whatever we multiply the denominators by. \[\begin{aligned} \frac{\sin{x}}{1+\cos{x}} + \frac{1+\cos{x}}{\sin{x}} &= \frac{2}{\sin{x}} \\ \frac{\sin{x}}{1+\cos{x}} \left( \frac{\sin{x}}{\sin{x}} \right) + \frac{1+\cos{x}}{\sin{x}} \left( \frac{1+\cos{x}}{1+\cos{x}} \right) &= \frac{2}{\sin{x}} \\ \frac{\sin^2{x}}{(1+\cos{x})\sin{x}} + \frac{(1+\cos{x})^2}{\sin{x}(1+\cos{x})} &= \frac{2}{\sin{x}} \\ \frac{\sin^2{x}+(1+\cos{x})^2}{\sin{x}(1+\cos{x})} &= \frac{2}{\sin{x}} \end{aligned}\] Then let’s foil the \((1+\cos{x})^2\) on the numerator. \[\begin{aligned} \frac{\sin^2{x}+1+2\cos{x}+\cos^2{x}}{\sin{x}(1+\cos{x})} &= \frac{2}{\sin{x}} \end{aligned}\] Now I see that I have a \(\sin^2{x}+\cos^2{x}\) on the numerator, so I can use the Pythagorean Identity \(\sin^2{x}+\cos^2{x}=1\). \[\begin{aligned} \frac{\sin^2{x}+1+2\cos{x}+\cos^2{x}}{\sin{x}(1+\cos{x})} &= \frac{2}{\sin{x}} \\ \frac{1+1+2\cos{x}}{\sin{x}(1+\cos{x})} &= \frac{2}{\sin{x}} \\ \frac{2+2\cos{x}}{\sin{x}(1+\cos{x})} &= \frac{2}{\sin{x}} \end{aligned}\] Now I can factor a 2 out of the numerator, and have a cancellation in the fraction. \[\begin{aligned} \frac{2(1+\cos{x})}{\sin{x}(1+\cos{x})} &= \frac{2}{\sin{x}} \\ \frac{2}{\sin{x}} &= \frac{2}{\sin{x}} \end{aligned}\]