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)
 
Right triangle2.png
 
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)
 
Right triangle2.png
 
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)
 
Right triangle2.png
 
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 \(90^\circ\).