Further trigonometric identities

​​In a nutshell

Using the trigonometric identity $$\sin^2(x)+\cos^2(x)=1$$, you can derive two new identities that involve reciprocal trigonometric functions.

Identities

Start with $$\sin^2(x)+\cos^2(x)=1$$ and divide both sides by $$\cos^2(x)$$:

​$$\begin{aligned}\sin^2(x)+\cos^2(x)&=1\\\frac{\sin^2(x)}{\cos^2(x)}+\frac{\cos^2(x)}{\cos^2(x)}&=\frac1{\cos^2(x)}\\\tan^2(x)+1&=\sec^2(x)\end{aligned}$$​​

Hence you have one new identity:

​$$\boxed{\tan^2(x)+1=\sec^2(x)}$$​​

Example 1

Prove that $$1+\cot^2(x)=\cosec^2(x)$$.

Again begin with $$\sin^2(x)+\cos^2(x)=1$$, but this time divide both sides by $$\sin^2(x)$$:

$$\begin{aligned}\sin^2(x)+\cos^2(x)&=1\\\frac{\sin^2(x)}{\sin^2(x)}+\frac{\cos^2(x)}{\sin^2(x)}&=\frac1{\sin^2(x)}\\\underline{1+\cot^2(x)\space}&\underline{=\cosec^2(x)}\end{aligned}$$​​

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Hence you have proven another identity:

​$$\boxed{\cot^2(x)+1=\cosec^2(x)}$$​​

Example 2

Solve the equation $$\cosec^2(x)+\cot(x)=3$$ in the interval $$0\leq x\leq360^\circ$$.

This equation involves $$\cosec(x)$$ and $$\cot(x)$$, so use the identity $$\cot^2(x)+1=\cosec^2(x)$$. You do this because solving this equation will be easier if the same trigonometric function is used throughout.

$$\begin{aligned}\cosec^2(x)+\cot(x)&=3\\(\cot^2(x)+1)+\cot(x)&=3\\\cot^2(x)+\cot(x)-2&=0\end{aligned}$$​​

Now you have a quadratic in $$\cot(x)$$. Factorising gives

$$\begin{aligned}\cot^2(x)+\cot(x)-2&=0\\(\cot(x)+2)(\cot(x)-1)&=0\end{aligned}$$​​

So $$\cot(x)=-2$$ or $$\cot(x)=1$$. Converting to equations in $$\tan(x)$$, you have $$\tan(x)=-0.5$$ or $$\tan(x)=1$$. Solving these gives

$$\underline{x = 45\degree, 153 \degree, 225 \degree, 333 \degree}$$​​

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