## Basic Identities

 6.4 Basic Identities

• There are three basic identities as follows:
 $$\text{sin}^2 \ \theta + \text{cos}^2 \ \theta = 1$$ $$1 + \text{tan}^2 \ \theta = \text{sec}^2 \ \theta$$ $$1 + \text{cot}^2 \ \theta = \text{cosec}^2 \ \theta$$

• All three basic identities can be derived by using a right-angled triangle $$ABC$$ and all the trigonometric ratios which have been learnt

 $$\text{sin }A = \dfrac{a}{c}$$ $$\text{cosec }A = \dfrac{c}{a}$$ $$\text{cos }A = \dfrac{b}{c}$$ $$\text{sec }A = \dfrac{c}{b}$$ $$\text{tan }A = \dfrac{a}{b}$$ $$\text{cot }A = \dfrac{b}{a}$$

• By using Pythagoras theorem, it is known that
 $$a^2+b^2 = c^2$$

• Divide the two sides of the equation by $$a^2, \ b^2 \text{ and }c^2$$, we get:

$$\div \ a^2$$ $$\div \ b^2$$ $$\div \ c^2$$
 \begin{aligned} &\dfrac{a^2}{a^2}+ \dfrac{b^2}{a^2} = \dfrac{c^2}{a^2}\\\\ &1 + \begin{pmatrix} \dfrac{b}{a} \end{pmatrix}^2 = \begin{pmatrix} \dfrac{c}{a} \end{pmatrix}^2\\\\ &1+ \text{cot}^2 \ A = \text{cosec}^2 \ A \end{aligned}
 \begin{aligned}& \dfrac{a^2}{b^2}+ \dfrac{b^2}{b^2} = \dfrac{c^2}{b^2}\\\\ &\begin{pmatrix} \dfrac{a}{b} \end{pmatrix}^2 + 1= \begin{pmatrix} \dfrac{c}{b} \end{pmatrix}^2\\\\ &1+ \text{tan}^2 \ A = \text{sec}^2 \ A \end{aligned}
 \begin{aligned} &\dfrac{a^2}{c^2}+ \dfrac{b^2}{c^2} = \dfrac{c^2}{c^2}\\\\ &\begin{pmatrix} \dfrac{a}{c} \end{pmatrix}^2+ \begin{pmatrix} \dfrac{b}{c} \end{pmatrix}^2 = 1\\\\ &\text{sin}^2 \ A+ \text{cos}^2 \ A = 1 \end{aligned}

• These three basic trigonometric identities can be used to solve problems involving trigonometric ratios

Example

Without using a calculator, find the value of each of the following:

 (a) $$\text{sin}^2 (-430^{\circ}) + \text{cos}^2(-430^{\circ})$$ (b) $$\text{tan}^2 \begin{pmatrix} \dfrac{\pi}{3} \end{pmatrix} - \text{sec}^2\begin{pmatrix} \dfrac{\pi}{3} \end{pmatrix}$$ Solution: (a) $$\text{sin}^2 \ A + \text{cos}^2 \ A=1\\ \text{sin}^2 (-430^{\circ}) + \text{cos}^2(-430^{\circ}) =1$$ (b) \begin{aligned} &1+ \text{tan}^2 \ A = \text{sec}^2 \ A\\\\ &\text{tan}^2 \begin{pmatrix} \dfrac{\pi}{3} \end{pmatrix} - \text{sec}^2\begin{pmatrix} \dfrac{\pi}{3} \end{pmatrix} =-1 \end{aligned}