Pythagorean Trigonometric Identities

Till 10th grade we are using pythagorean trigonometric identities but from 11th grade we will prove this identity. Pythagorean identities on Trigonometry are as follows :

1) $sin^{2} \Theta + cos^{2} \Theta$ = 1

2) 1 + $tan^{2} \Theta = sec^{2} \Theta$

3) 1 + $cot^{2} \Theta = csc^{2} \Theta$
Proof : According to the diagram,

1) In triangle OMP we have, $sin \Theta = \frac{PM}{OP}$

∴ $sin^{2} \Theta = \left ( \frac{PM}{OP} \right )^{2}$ ------(i)

$cos \Theta = \frac{OM}{OP}$

∴ $cos^{2} \Theta = \left ( \frac{OM}{OP} \right )^{2}$-------(ii)

Adding (i) and (ii), we get
$sin^{2} \Theta + cos^{2} \Theta = \left ( \frac{PM}{OP} \right )^{2} + \left ( \frac{OM}{OP} \right )^{2}$

                = $\frac{PM^{2}}{OP^{2}} + \frac{OM^{2}}{OP^{2}}$

                = $\frac{PM^{2} + OM^{2}}{OP^{2}}$
By pythagorean theorem in $\Delta$ OMP, $ PM^{2} + OM^{2} = OP^{2}$
$sin^{2} \Theta + cos^{2} \Theta = \frac{OP^{2}}{OP^{2}}$

$sin^{2} \Theta + cos^{2} \Theta$ = 1

2) In triangle OMP we have, $tan \Theta = \frac{PM}{OM}$

∴ $tan^{2} \Theta = \left ( \frac{PM}{OM} \right )^{2}$ ------(i)

$sec \Theta = \frac{OP}{OM}$

∴ $sec^{2} \Theta = \left ( \frac{OP}{OM} \right )^{2}$-------(ii)

1 + $tan^{2} \Theta = 1 + \left ( \frac{PM}{OM} \right )^{2}$

                = 1 + $\frac{PM^{2}}{OM^{2}}$

                = $\frac{OM^{2} + PM^{2}}{OM^{2}}$
By pythagorean theorem in $\Delta$ OMP, $ PM^{2} + OM^{2} = OP^{2}$
1 + $tan^{2} \Theta = \frac{OP^{2}}{OM^{2}}$

1 + $tan^{2} \Theta = sec^{2} \Theta$

3) In triangle OMP we have,
$cot \Theta = \frac{OM}{PM}$

∴ $cot^{2} \Theta = \left ( \frac{OM}{PM} \right )^{2}$ ------(i)

$csc \Theta = \frac{OP}{PM}$

∴ $csc^{2} \Theta = \left ( \frac{OP}{PM} \right )^{2}$-------(ii)

1 + $cot^{2} \Theta = 1 + \left ( \frac{OM}{PM} \right )^{2}$

                = 1 + $\frac{OM^{2}}{PM^{2}}$

                = $\frac{PM^{2} + OM^{2}}{PM^{2}}$
By pythagorean theorem in $\Delta$ OMP, $ PM^{2} + OM^{2} = OP^{2}$
1 + $cot^{2} \Theta = \frac{OP^{2}}{PM^{2}}$

1 + $cot^{2} \Theta = csc^{2} \Theta$

More reults on Pythagorean trigonometric identities

1) $sin^{2} \Theta = 1 - cos^{2} \Theta$

2) $cos^{2} \Theta = 1 - sin^{2} \Theta$

3) $tan^{2} \Theta = sec^{2} \Theta$ - 1

4) $sec^{2} \Theta - tan^{2} \Theta$ = 1

5) $tan^{2} \Theta - sec^{2} \Theta$ = - 1

6) $cot^{2} \Theta = csc^{2} \Theta$ - 1

7) $csc^{2} \Theta - cot^{2} \Theta$ = 1

8) $cot^{2} \Theta - csc^{2} \Theta$ = - 1

11th grade math

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