Trigonometric Proofs Exercise 2

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.