UPSKILL MATH PLUS
Learn Mathematics through our AI based learning portal with the support of our Academic Experts!
Learn moreA 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 \(90^\circ\).