### Theory:

A trigonometric identity is an equation containing trigonometric ratios of an angle that is true for all values of the angle(s) involved.
Fundamental identities of trigonometry:

(1) $$\sin^2 \theta + \cos^2 \theta = 1$$

(2) $$1 + \tan^2 \theta = \sec^2 \theta$$

(3) $$1 + \cot^2 \theta = \text{cosec}^2\: \theta$$
We will prove these basic trigonometric identities.
$$\sin^2 \theta + \cos^2 \theta = 1$$
Consider a right triangle $$ABC$$ with angle $$\theta$$.

$$\sin \theta = \frac{AB}{AC}$$ - - - - - (I)

$$\cos \theta = \frac{BC}{AC}$$ - - - - - (II) By Pythagoras theorem:

$$AB^2 + BC^2 = AC^2$$

Divide by $$AC^2$$ on both sides.

$$\frac{AB^2}{AC^2} + \frac{BC^2}{AC^2} = \frac{AC^2}{AC^2}$$

$$\left(\frac{AB}{AC}\right)^2 + \left(\frac{BC}{AC}\right)^2 = \left(\frac{AC}{AC}\right)^2$$

$$(\sin \theta)^2 + (\cos \theta)^2 = 1^2$$ [using equation (I) and (II)]

$$\sin^2 \theta + \cos^2 \theta = 1$$

Hence, the identity is $$\sin^2 \theta + \cos^2 \theta = 1$$.
$$1 + \tan^2 \theta = \sec^2 \theta$$
Consider a right triangle $$ABC$$ with angle $$\theta$$.

$$\tan \theta = \frac{AB}{BC}$$ - - - - - (III)

$$\sec \theta = \frac{AC}{BC}$$ - - - - - (IV) By Pythagoras theorem:

$$AB^2 + BC^2 = AC^2$$

Divide by $$BC^2$$ on both sides.

$$\frac{AB^2}{BC^2} + \frac{BC^2}{BC^2} = \frac{AC^2}{BC^2}$$

$$\left(\frac{AB}{BC}\right)^2 + \left(\frac{BC}{BC}\right)^2 = \left(\frac{AC}{BC}\right)^2$$

$$(\tan \theta)^2 + 1^2 = (\sec \theta)^2$$ [using equation (III) and (IV)]

$$\tan^2 \theta + 1 = \sec^2 \theta$$

Hence, the identity is $$1 + \tan^2 \theta = \sec^2 \theta$$.
$$1 + \cot^2 \theta = \text{cosec}^2\: \theta$$
Consider a right triangle $$ABC$$ with angle $$\theta$$.

$$\cot \theta = \frac{BC}{AB}$$ - - - - - (III)

$$\text{cosec}\: \theta = \frac{AC}{AB}$$ - - - - - (IV) By Pythagoras theorem:

$$AB^2 + BC^2 = AC^2$$

Divide by $$AB^2$$ on both sides.

$$\frac{AB^2}{AB^2} + \frac{BC^2}{AB^2} = \frac{AC^2}{AB^2}$$

$$\left(\frac{AB}{AB}\right)^2 + \left(\frac{BC}{AB}\right)^2 = \left(\frac{AC}{AB}\right)^2$$

$$1^2 + (\cot \theta)^2 = (\text{cosec}\: \theta)^2$$ [using equation (III) and (IV)]

$$1 + \cot^2 \theta + 1 = \text{cosec}^2\: \theta$$

Hence, the identity is $$1 + \cot^2 \theta + 1 = \text{cosec}^2\: \theta$$.
Equal forms of trigonometric identities
 Identity Equal form of identity $$\sin^2 \theta + \cos^2 \theta = 1$$ $$\sin^2 \theta = 1 - \cos^2 \theta$$ (or) $$\cos^2 \theta = 1 - \sin^2 \theta$$ $$1 + \tan^2 \theta = \sec^2 \theta$$ $$\tan^2 \theta = \sec^2 \theta - 1$$ (or) $$\sec^2 \theta - \tan^2 \theta = 1$$ $$1 + \cot^2 \theta = \text{cosec}^2\: \theta$$ $$\cot^2 \theta = \text{cosec}^2\: \theta - 1$$ (or) $$\text{cosec}^2\: \theta - \cot^2 \theta = 1$$

Important!
The three trigonometric identities are true for every $$\theta$$ lies between $$0^\circ$$ and $$90^\circ$$.