To prove a new trig identity, we need to show that one side is equivalent to the other using established identities, in this case, the Big 8 (the Ratio, Reciprocal and Pythagorean Identities). You only want to work on one side, so pick the more complicated one and try to simplify it. Looking at this one, the more complicated side looks like the LHS, so I’m going to try manipulating the LHS to make it look like the RHS.

First, I’m going to FOIL out the \((1+\cot{x})^2\), \[\begin{aligned} (1+\cot{x})^2\sin^2{x}&=1+2\sin{x}\cos{x} \\ (1+2\cot{x}+\cot^2{x})\sin^2{x} &= 1+2\sin{x}\cos{x} \end{aligned}\] Then I notice that the RHS is in terms of cosine and sine so I’ll convert the LHS to terms of sine and cosine. So I need to use the ratio identity \(\cot{x}=\frac{\cos{x}}{\sin{x}}\), \[\begin{aligned} (1+2\cot{x}+\cot^2{x})\sin^2{x} &= 1+2\sin{x}\cos{x} \\ (1+2\frac{\cos{x}}{\sin{x}}+\frac{\cos^2{x}}{\sin^2{x}})\sin^2{x} &= 1+2\sin{x}\cos{x} \end{aligned}\] Then I can distribute the \(\sin^2{x}\), \[\begin{aligned} (1+2\frac{\cos{x}}{\sin{x}}+\frac{\cos^2{x}}{\sin^2{x}})\sin^2{x} &= 1+2\sin{x}\cos{x} \\ \sin^2{x}+2\frac{\cos{x} \cdot \sin^2{x}}{\sin{x}}+\frac{\cos^2{x} \cdot \sin^2{x}}{\sin^2{x}} &= 1+2\sin{x}\cos{x} \\ \sin^2{x} +2\sin{x}\cos{x}+\cos^2{x} &= 1+2\sin{x}\cos{x} \end{aligned}\] Now I have a \(\sin^2{x} \text{ and } \cos^2{x}\), so I can use the Pythagorean Identity \(\sin^2{x}+\cos^2{x}=1\) \[\begin{aligned} \sin^2{x} +2\sin{x}\cos{x}+\cos^2{x} &= 1+2\sin{x}\cos{x} \\ 1+2\sin{x}\cos{x} &= 1+2\sin{x}\cos{x} \end{aligned}\]

A helpful strategy in proving new trig identities is factoring out common terms. We can see that the LHS of this equation is the more complicated side that we want to work with. I also see I have a common factor in both terms, \(\cot{x}\). So let’s factor out a \(\cot{x}\) from the LHS. \[\begin{aligned} \sec^2{x}\cot{x}-\cot{x} &= \tan{x} \\ \cot{x}(\sec^2{x}-1) &= \tan{x} \end{aligned}\] Now I have a factor \((\sec^2{x}-1)\). I recall there is a Pythagorean identity with a \(\sec^2{x}\) and a 1: \[1+\tan^2{x} = \sec^2{x}\] So I can rearrange the identity to the form of the factor, \[\tan^2{x} = \sec^2{x}-1\] So now I can substitute in \(\tan^2{x}\), \[\begin{aligned} \cot{x}(\tan^2{x}) &= \tan{x} \end{aligned}\] Then I know the ratio identity \[\cot{x} = \frac{1}{\tan{x}}\] So I have \[\begin{aligned} \frac{1}{\tan{x}}(\tan^2{x}) &= \tan{x} \\ \tan{x} &= \tan{x} \end{aligned}\] And I’ve proven the identity.

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. \[\begin{aligned} \frac{2(1+\cos{x})}{\sin{x}(1+\cos{x})} &= \frac{2}{\sin{x}} \\ \frac{2}{\sin{x}} &= \frac{2}{\sin{x}} \end{aligned}\]

When the numerator contains terms that are also factors of the denominator, splitting a fraction can be an effective first step in proving a trig identity. In this identity, I can see a cotangent and a cosine in the numerator, both of which are factors of the denominator. So I’m going to split this fraction with a difference in the numerator, into a difference of two fractions with a common denominator. (Remember, you can only split a fraction with a sum/difference in the numerator. You can’t split a denominator.) \[\begin{aligned} \frac{\cot{x}-\cos{x}\sin{x}}{\cot{x}\cos{x}} &= \cos{x} \\ \frac{\cot{x}}{\cot{x}\cos{x}} - \frac{\cos{x}\sin{x}}{\cot{x}\cos{x}} &= \cos{x} \\ \frac{1}{\cos{x}} - \frac{\sin{x}}{\cot{x}} &= \cos{x} \end{aligned}\]

Then we see that everything but that cotangent is in terms of sine and cosine, so I’ll use the ratio identity \(\cot{x}=\frac{\cos{x}}{\sin{x}}\). \[\begin{aligned} \frac{1}{\cos{x}} - \frac{\sin{x}}{\frac{\cos{x}}{\sin{x}}} &= \cos{x} \end{aligned}\] So we copy-dot-flip, \[\begin{aligned} \frac{1}{\cos{x}} - \sin{x} \left( \frac{\sin{x}}{\cos{x}} \right) &= \cos{x} \\ \frac{1}{\cos{x}} - \frac{\sin^2{x}}{\cos{x}} &= \cos{x} \end{aligned}\] Now we have a common denominator again so I’ll combine these fractions, \[\begin{aligned} \frac{1-\sin^2{x}}{\cos{x}} &= \cos{x} \end{aligned}\] And now the numerator is a rearrangement of the Pythagorean identity \[\begin{aligned} \sin^2{x}+\cos^2{x}=1 \\ \cos^2{x} = 1 - \sin^2{x} \end{aligned}\] So we plug that in and simplify. \[\begin{aligned} \frac{\cos^2{x}}{\cos{x}} &= \cos{x} \\ \cos{x} &= \cos{x} \end{aligned}\]