One Right Triangle Hides Three Identities That Hold for Every Angle

An identity is an equation that is true for every value of the angle for which both sides are defined — unlike an ordinary equation, which is true only for particular values.

In a right triangle with hypotenuse $c$ and the other sides $a$ opposite $A$ and $b$ adjacent to it, Pythagoras gives $a^2+b^2=c^2$. **Dividing by $c^2$, $b^2$ or $a^2$** produces the three basic identities.

This part covers each identity, proving identities, and the values at standard angles.

**Dividing $a^2+b^2=c^2$ by $c^2$ gives ${\sin }^{2}A+{\cos }^{2}A=1$, which lets you replace $1-{\cos }^{2}A$ with ${\sin }^{2}A$ or $1-{\sin }^{2}A$ with ${\cos }^{2}A$.

Worked example 1.** Simplify $\frac{1-{\cos }^{2}A}{\sin A}$.

$$\frac{{\sin }^{2}A}{\sin A}=\sin A$$

Worked example 2. Simplify ${\sin }^{4}A-{\cos }^{4}A$.

$$({\sin }^{2}A-{\cos }^{2}A)({\sin }^{2}A+{\cos }^{2}A)={\sin }^{2}A-{\cos }^{2}A$$

Worked example 3. Show $(\sin A+\cos A{)}^2+(\sin A-\cos A{)}^2=2$.

$$2{\sin }^{2}A+2{\cos }^{2}A=2({\sin }^{2}A+{\cos }^{2}A)=2$$

Worked example 4. If $\sin A=\frac{3}{5}$ and $A$ is acute:

$$\cos A=\sqrt{1-\frac{9}{25}}=\frac{4}{5}\tan A=\frac{3}{4}$$

An everyday example. **A $5$ m ladder leaning against a wall** reaches height $5\sin A$ and stands $5\cos A$ from the wall, and $(5\sin A{)}^2+(5\cos A{)}^2=25$ is just Pythagoras.

The substance. **${\sin }^{2}A$ means $(\sin A{)}^2$**, not the sine of $A^2$.

**Dividing ${\sin }^{2}A+{\cos }^{2}A=1$ by ${\cos }^{2}A$ gives $1+{\tan }^{2}A={\sec }^{2}A$, and dividing by ${\sin }^{2}A$ gives $1+{\cot }^{2}A={\mathrm{cosec}}^{2}A$.**

Useful rearrangements: ${\sec }^{2}A-{\tan }^{2}A=1$ and ${\mathrm{cosec}}^{2}A-{\cot }^{2}A=1$.

Worked example 1. Simplify $({\sec }^{2}A-1){\cot }^{2}A$.

$${\tan }^{2}A\times {\cot }^{2}A=1$$

Worked example 2. Simplify $(1+{\cot }^{2}A){\sin }^{2}A$.

$${\mathrm{cosec}}^{2}A\times {\sin }^{2}A=1$$

Worked example 3. Show ${\sec }^{4}A-{\sec }^{2}A={\tan }^{4}A+{\tan }^{2}A$.

$${\sec }^{2}A({\sec }^{2}A-1)=(1+{\tan }^{2}A){\tan }^{2}A={\tan }^{2}A+{\tan }^{4}A$$

Worked example 4. If $\tan A=\frac{5}{12}$, then $\sec A=\sqrt{1+\frac{25}{144}}=\frac{13}{12}$ for acute $A$.

An everyday example. **A wheelchair ramp rising $5$ m over $12$ m** has $\tan A=\frac{5}{12}$, and its sloping length is $\sec A=\frac{13}{12}$ times the horizontal run.

The substance. **${\sec }^{2}A-{\tan }^{2}A=1$ factorises as $(\sec A+\tan A)(\sec A-\tan A)=1$**, a trick used in Part 2.

**Start with the more complicated side, replace tan, cot, sec and cosec by their sine and cosine forms, combine fractions, and use ${\sin }^{2}A+{\cos }^{2}A=1$ until it matches the other side.

Worked example 1.** Prove $\tan A+\cot A=\sec A\mathrm{cosec}A$.

$$\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}=\frac{{\sin }^{2}A+{\cos }^{2}A}{\sin A\cos A}=\frac{1}{\sin A\cos A}=\sec A\mathrm{cosec}A$$

Worked example 2. Prove $(1-\sin A)(\sec A+\tan A)=\cos A$.

$$(1-\sin A)\times \frac{1+\sin A}{\cos A}=\frac{1-{\sin }^{2}A}{\cos A}=\frac{{\cos }^{2}A}{\cos A}=\cos A$$

Worked example 3. Prove $\mathrm{cosec}A-\sin A=\cot A\cos A$.

$$\frac{1}{\sin A}-\sin A=\frac{1-{\sin }^{2}A}{\sin A}=\frac{{\cos }^{2}A}{\sin A}=\frac{\cos A}{\sin A}\times \cos A$$

An everyday example. Just as you convert rupees and paise to the same unit before adding a bill, convert every ratio to sine and cosine before combining.

The substance. Work on one side only; an identity is not an equation, so cross-multiplying both sides is not a valid proof.

**Substitute the exact values of the ratios, simplify fractions and surds, and remember that $\tan 90^{\circ }$ and $\sec 90^{\circ }$ are not defined.

Key values:**

- $\sin$: $0,\frac{1}{2},\frac{1}{\sqrt{2}},\frac{\sqrt{3}}{2},1$
- $\cos$: $1,\frac{\sqrt{3}}{2},\frac{1}{\sqrt{2}},\frac{1}{2},0$
- $\tan$: $0,\frac{1}{\sqrt{3}},1,\sqrt{3}$, not defined

Worked example 1.

$$\sin 30^{\circ }\cos 60^{\circ }+\cos 30^{\circ }\sin 60^{\circ }=\frac{1}{4}+\frac{3}{4}=1$$

Worked example 2.

$$\frac{{\tan }^{2}60^{\circ }+4{\cos }^{2}45^{\circ }+3{\sec }^{2}30^{\circ }+5{\cos }^{2}90^{\circ }}{\mathrm{cosec}30^{\circ }+\sec 60^{\circ }-{\cot }^{2}30^{\circ }}=\frac{3+2+4+0}{2+2-3}=9$$

Worked example 3 — checking the identity. ${\sin }^{2}45^{\circ }+{\cos }^{2}45^{\circ }=\frac{1}{2}+\frac{1}{2}=1$.

An everyday example. **A roof sloping at $30^{\circ }$** rises $\tan 30^{\circ }\approx 0.577$ m for every metre across.

The substance. **${\sec }^{2}30^{\circ }=\frac{4}{3}$**, so $3{\sec }^{2}30^{\circ }=4$ — square the ratio before multiplying.