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tags:
Fundamental Trigonometric Identity Trigonometry Trigonometry Series

Trigonometric Proofs Exercise 2

Author: Hannah Solomon

The following problem is solved in this video. It is recommended that you try to solve the problem before watching the video. You can click "Reveal Answer" to see the answer to the problem.

Problem: Prove the following identity.
\[\sec^2{x}\cot{x}-\cot{x} = \tan{x}\]

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.

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