# Trigonometric Identities

Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. An identity in equalities which are true for every value occurring on both sides of an equation. Geometrically, these identities involving certain functions of one or more angles. There are various distinct identities involving the side length as well as the angle of a triangle.

## Trigonometric Identities

Similarly, an equation which involves trigonometric ratios of an angle represents a trigonometric identity.
The upcoming discussion covers the fundamental trigonometric identities and their proofs.
Consider a right angle $∆ABC$ which is right-angled at $B$ as shown in the given figure.

Applying Pythagoras Theorem for the given triangle, we have

$(hypotenuse)^2$ = $(base)^2 + (perpendicular)^2$

$\mathbf{AC^2 = AB^2+BC^2}$………………………..(1)

### Identity 1

Now, divide each term of equation (1) by $AC^2$, we have

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

$⇒\frac{AB^2}{AC^2}+\frac{BC^2}{AC^2}$ = $1$

$⇒(\frac{AB}{AC})^2+(\frac{BC}{AC})^2$ = $1$ ………………………….(2)

We know that,
$(\frac{AB}{AC})^2 = \cos a$ and $(\frac{BC}{AC})^2 = \sin a$, thus equation (2) can be written as-

$\mathbf{\sin^{2}a + \cos ^{2}a = 1}$ (IDENTITY 1)

Identity 1 is valid for angles $0 \leq a \leq 90$

### Identity 2

Now Dividing the equation (1) by $AB^2$, we get

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

$⇒\frac{AC^2}{AB^2}$ = $1+\frac{BC^2}{AB^2}$

$⇒ (\frac{AC}{AB})^2$ = $1+(\frac{BC}{AB})^2$ ……………………(3)

By referring trigonometric ratios, it can be seen that:

$\frac{AC}{AB}$ = $\frac{hypotenuse}{side~ adjacent~ to ~angle~ a}$ = $\sec a$

Similarly,

$\frac{BC}{AB}$ = $\frac{side~ opposite~ to~ angle~ a}{side ~adjacent~ to~ angle~ a}$ = $\tan a$

Replacing the values of $\frac{AC}{AB}$ and $\frac{BC}{AB}$ in the equation (3) gives,

$\mathbf{1+tan^2 a = \sec^2 a}$ (IDENTITY 2)

As it is known that $\tan a$ is not defined for $a$ = 90° therefore  identity 2 obtained above is true for 0 ≤ A <90.

### Identity 3

Dividing the equation (1) by $BC^2$, we get

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

$⇒\frac{AC^2}{BC^2}$ = $\frac{AB^2}{BC^2}+1$

$⇒(\frac{AC}{BC})^2$ = $(\frac{AB}{BC})^2+1$ …………………..(iv)

By referring trigonometric ratios, it can be seen that:

$\frac{AC}{BC}$ = $\frac{hypotenuse}{side ~opposite~ to~ angle~ a}$ = $cosec a$

Also,

$\frac{AB}{BC}$ = $\frac{side~ adjacent~ to~ angle~ a}{side~ opposite ~to ~angle~ a}$ = $\cot a$

Replacing the values of $\frac{AC}{BC}$ and $\frac{AB}{BC}$ in the equation (4) gives,

$\mathbf{cosec~^2~a = 1~+~cot^2~a}$ (IDENTITY 3)

Since $cosec~a$ and $\cot~a$ are not defined for $a$ = 0° therefore the identity 3 is obtained is true for all the values of $a$ except at $a$ = 0°. Therefore, the identity is true for all $a$ such that, 0° < $a$ ≤ 90°.

Thus, the three fundamental trigonometric identities can be listed as:

• $sin^2~a~+~cos^2~a$ = $1$
• $1~+~tan^2~a$ = $sec^2~a$
• $1~+~cot^2~a$ = $cosec^2~a$

Example: Consider a $∆ABC$, right angled at $B$. The length of base, $AB$ = 4 cm and length of perpendicular

$BC$ =3 cm. Find the value of $sec~A$.

As the length of perpendicular and base is given; it can be concluded that,

$tan~A$ = $\frac{3}{4}$

Now, using the trigonometric identity $1~+~tan^2~A$ = $sec^2~A$

$sec^2~A$ = $1~+~(\frac{3}{4})^2$

$sec^2~A$ = $\frac{25}{16}$

$⇒sec~A$ = $±\frac{5}{4}$

Since, the ratio of lengths is positive, we can neglect $sec~A$ = $-\frac{5}{4}$.

Therefore, $sec~A$ = $\frac{5}{4}$

Angle Sum and Difference Identities-

Consider a Triangle $\alpha$ and $\beta$, the trigonometric sum and differenece identities are as follows-

$\sin (\alpha + \beta) = \sin (\alpha) . \cos (\beta) + \cos (\alpha ). \sin (\beta)$

$\sin (\alpha – \beta) = \sin \alpha . \cos \beta – \cos \alpha . \sin \beta$

$\cos (\alpha + \beta) = \cos \alpha . \cos \beta – \sin \alpha . \sin \beta$

$\cos (\alpha – \beta) = \cos \alpha . \cos \beta + \sin \alpha . \sin \beta$

$\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 – \tan \alpha. \tan \beta}$

$\tan (\alpha – \beta) = \frac{\tan \alpha – \tan \beta}{1 + \tan \alpha. \tan \beta}$

Using these identities, we can solve various mathematical problems. All you need to know about trigonometry and its applications is just a click away,  visit BYJU’S to learn more.

#### Practise This Question

Simplify the expression-
(cosec2θ1)×(sec2θ1)