Subtraction of square of tangent from square of secant at an angle equals to $1$ is called Pythagorean trigonometric identity of secant and tangent

The relation of secant with tangent is derived by using Pythagorean Theorem. Hence, it is called as Pythagorean trigonometric identity and also called as Pythagoras trigonometric identity in trigonometry. It is mainly used to express secant in terms of tangent and also used to express tangent in terms of secant.

$\Delta BAC$ is a right angled triangle and angle of the triangle is theta $(\theta)$.

Apply Pythagoras Theorem to this triangle.

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

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

Divide left hand side and right hand side by ${AB}^2$.

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

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

$$\Rightarrow \Bigg(\frac{BC}{AB}\Bigg)^2 + \Bigg(\frac{AB}{AB}\Bigg)^2 = \Bigg(\frac{AC}{AB}\Bigg)^2$$

$$\Rightarrow \Bigg(\frac{BC}{AB}\Bigg)^2 + 1 = \Bigg(\frac{AC}{AB}\Bigg)^2$$

According to right angled triangle $BAC$

$$\frac{BC}{AB} = \tan \theta$$

$$\frac{AC}{AB} = \sec \theta$$

Substitute these two ratios in terms of associated trigonometric functions.

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

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

$$\Rightarrow 1 = \sec^2 \theta \, – \, \tan^2 \theta$$

$$\Rightarrow \sec^2 \theta \, – \, \tan^2 \theta = 1$$

It is proved that secant squared angle minus tangent squared angle is equal $1$.

Assume angle of triangle $\theta = 45^°$ to verify this trigonometric identity mathematically.

$$\sec^2 \theta \, – \, \tan^2 \theta = \sec^2 45^° \, – \, \tan^2 45^°$$

$$\sec^2 45^° \, – \, \tan^2 45^° = (\sqrt{2})^2 \,- (1)^2$$

$$\sec^2 45^° \, – \, \tan^2 45^° = 2 \,- 1$$

$$\sec^2 45^° \, – \, \tan^2 45^° = 1$$

It has proved that subtraction of square of tangent from square of secant at the angle $45^°$ is $1$. It is also proved in case of every angle. Hence, this mathematical relation between secant and tangent is called as a trigonometric identity in trigonometry.

The Pythagoras trigonometric identity of tangent and secant can also be used to write secant in terms of tangent and also helpful to express tangent in terms of secant.

1

Secant can be written in terms of tangent in two different forms.

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

$$\sec \theta = \pm \sqrt{1 + \tan^2 \theta}$$

2

Tangent can be written in terms of secant in two different forms.

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

$$\tan \theta = \pm \sqrt{\sec^2 \theta \,-\, 1}$$

Copyright © 2012 - 2017 Math Doubts, All Rights Reserved