AHL 3.9—Reciprocal trig ratios and their pythagorean identities. Inverse circular functions

You already know sine, cosine, and tangent , but there are three more trigonometric functions built directly from their reciprocals. These are cosecant, secant, and cotangent, and they appear frequently in calculus, integration, and identities at HL.

Cosecant (csc): The reciprocal of sine: $\csc \theta =\frac{1}{\sin \theta }=\frac{\text{hypotenuse}}{\text{opposite}}$

Secant (sec): The reciprocal of cosine: $\sec \theta =\frac{1}{\cos \theta }=\frac{\text{hypotenuse}}{\text{adjacent}}$

Cotangent (cot): The reciprocal of tangent: $\cot \theta =\frac{1}{\tan \theta }=\frac{\cos \theta }{\sin \theta }=\frac{\text{adjacent}}{\text{opposite}}$

Note that cotangent can also be written as $\frac{\cos \theta }{\sin \theta }$, which is useful when simplifying expressions.

You already know the fundamental Pythagorean identity ${\sin }^2\theta +{\cos }^2\theta =1$, derived by dividing the right-triangle relation $o^2+a^2=h^2$ by $h^2$. By dividing by $o^2$ or $a^2$ instead, we obtain two additional identities involving the reciprocal functions.

Dividing $o^2+a^2=h^2$ by $o^2$:
$1+\frac{a^2}{o^2}=\frac{h^2}{o^2}$
Since $\cot \theta =\frac{a}{o}$ and $\csc \theta =\frac{h}{o}$, this becomes:
$\boxed{1+{\cot }^2\theta ={\csc }^2\theta }$

Dividing $o^2+a^2=h^2$ by $a^2$:
$\frac{o^2}{a^2}+1=\frac{h^2}{a^2}$
Since $\tan \theta =\frac{o}{a}$ and $\sec \theta =\frac{h}{a}$, this becomes:
$\boxed{{\tan }^2\theta +1={\sec }^2\theta }$

The three Pythagorean identities together are:

1. ${\sin }^2\theta +{\cos }^2\theta =1$
2. $1+{\cot }^2\theta ={\csc }^2\theta$
3. ${\tan }^2\theta +1={\sec }^2\theta$