(\boxed{\tan{\theta}≡\dfrac{\sin{\theta}}{\cos{\theta}}})

Applying the reciprocal function: (\cot{\theta}≡\dfrac{1}{\tan{\theta}})

(\implies \boxed{\cot{\theta}≡\dfrac{\cos{\theta}}{\sin{\theta}}})

Consider the right angle triangle with hypotenuse of length (1):

mtaimg/images/topics/5/5-55-1.png/mtaimg

(\sin{\theta}≡\dfrac{O}{1} \quad \cos{\theta}≡\dfrac{A}{1})

Applying Pythagoras' Theorem: (O^2+A2=12)

(\implies \boxed{\sin^{2{\theta}+\cos2{\theta}} = 1})

Dividing by (\cos^2{\theta}): (\dfrac{\sin^2{\theta}}{\cos^2{\theta}}+\dfrac{\cos^2{\theta}}{\cos^2{\theta}} = \dfrac{1}{\cos^2{\theta}})

(\implies \boxed{\tan^2{\theta}+1 = \sec^2{\theta}})

Dividing by (\sin^2{\theta}): (\dfrac{\sin^2{\theta}}{\sin^2{\theta}}+\dfrac{\cos^2{\theta}}{\sin^2{\theta}} = \dfrac{1}{\sin^2{\theta}})

(\implies \boxed{1+\cot^2{\theta} = \cosec^2{\theta}})

These identities can be used to solve trigonometric equations and to prove further identities.

[b]uTrigonometric identities[/u]/b

(\tan{\theta}≡\dfrac{\sin{\theta}}{\cos{\theta}})

(\cot{\theta}≡\dfrac{\cos{\theta}}{\sin{\theta}})

(\sin^{2{\theta}+\cos2{\theta}} = 1)

(\tan^2{\theta}+1 = \sec^2{\theta})

(1+\cot^2{\theta} = \cosec^2{\theta})