$$(tan A - tan B)/(1 + tan A tan B)$$

$$(tan A + tan B)/(1 + tan A tan B)$$

$$(tan A + tan B)/(1 - tan A tan B)$$

$$(1 + tan A tan B)/(tan A - tan B)$$

$$-(cos x cos y + sin x sin y)$$

Trigonometric Identities (List of Trigonometric Identities | Proofs

Trigonometric Identities are useful whenever trigonometric functions are involved in an expression or an equation. Trigonometric Identities are true for every value of variables occurring on both sides of an equation. Geometrically, these identities involve certain trigonometric functions (such as sine, cosine, tangent) of one or more angles.

Sine, cosine and tangent are the primary trigonometry functions whereas cotangent, secant and cosecant are the other three functions. The trigonometric identities are based on all the six trig functions. Check Trigonometry Formulas to get formulas related to trigonometry.

Table of Contents:

Definition
List of Trig Functions
Proofs
Triangle Identities
Solved Examples
Practice Questions
FAQs

What are Trigonometric Identities?

Trigonometric Identities are the equalities that involve trigonometry functions and holds true for all the values of variables given in the equation.

There are various distinct trigonometric identities involving the side length as well as the angle of a triangle. The trigonometric identities hold true only for the right-angle triangle.

All the trigonometric identities are based on the six trigonometric ratios. They are sine, cosine, tangent, cosecant, secant, and cotangent. All these trigonometric ratios are defined using the sides of the right triangle, such as an adjacent side, opposite side, and hypotenuse side. All the fundamental trigonometric identities are derived from the six trigonometric ratios.

Trigonometric Identities PDF

Click here to download the PDF of trigonometry identities of all functions such as sin, cos, tan and so on.

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List of Trigonometric Identities

There are various identities in trigonometry which are used to solve many trigonometric problems. Using these trigonometric identities or formulas, complex trigonometric questions can be solved quickly. Let us see all the fundamental trigonometric identities here.

Reciprocal Trigonometric Identities

The reciprocal trigonometric identities are:

Sin θ = 1/Csc θ or Csc θ = 1/Sin θ
Cos θ = 1/Sec θ or Sec θ = 1/Cos θ
Tan θ = 1/Cot θ or Cot θ = 1/Tan θ

Pythagorean Trigonometric Identities

There are three Pythagorean trigonometric identities in trigonometry that are based on the right-triangle theorem or Pythagoras theorem.

sin²a + cos²a = 1
1+tan² a = sec² a
cosec²a = 1 + cot²a

Ratio Trigonometric Identities

The trigonometric ratio identities are:

Tan θ = Sin θ/Cos θ
Cot θ = Cos θ/Sin θ

Trigonometric Identities of Opposite Angles

The list of opposite angle trigonometric identities are:

Sin (-θ) = – Sin θ
Cos (-θ) = Cos θ
Tan (-θ) = – Tan θ
Cot (-θ) = – Cot θ
Sec (-θ) = Sec θ
Csc (-θ) = -Csc θ

Trigonometric Identities of Complementary Angles

In geometry, two angles are complementary if their sum is equal to 90 degrees. Similarly, when we can learn here the trigonometric identities for complementary angles.

Sin (90 – θ) = Cos θ
Cos (90 – θ) = Sin θ
Tan (90 – θ) = Cot θ
Cot ( 90 – θ) = Tan θ
Sec (90 – θ) = Csc θ
Csc (90 – θ) = Sec θ

Trigonometric Identities of Supplementary Angles

Two angles are supplementary if their sum is equal to 90 degrees. Similarly, when we can learn here the trigonometric identities for supplementary angles.

sin (180°- θ) = sinθ
cos (180°- θ) = -cos θ
cosec (180°- θ) = cosec θ
sec (180°- θ)= -sec θ
tan (180°- θ) = -tan θ
cot (180°- θ) = -cot θ

Sum and Difference of Angles Trigonometric Identities

Consider two angles , α and β, the trigonometric sum and difference identities are as follows:

sin(α+β)=sin(α).cos(β)+cos(α).sin(β)
sin(α–β)=sinα.cosβ–cosα.sinβ
cos(α+β)=cosα.cosβ–sinα.sinβ
cos(α–β)=cosα.cosβ+sinα.sinβ
$\begin{array}{l}\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 – \tan \alpha. \tan \beta}\end{array} $
$\begin{array}{l}\tan (\alpha – \beta) = \frac{\tan \alpha – \tan \beta}{1 + \tan \alpha. \tan \beta}\end{array} $

Double Angle Trigonometric Identities

If the angles are doubled, then the trigonometric identities for sin, cos and tan are:

sin 2θ = 2 sinθ cosθ
cos 2θ = cos²θ – sin²θ = 2 cos²θ – 1 = 1 – 2sin² θ
tan 2θ = (2tanθ)/(1 – tan²θ)

Half Angle Identities

If the angles are halved, then the trigonometric identities for sin, cos and tan are:

sin (θ/2) = ±√[(1 – cosθ)/2]
cos (θ/2) = ±√(1 + cosθ)/2
tan (θ/2) = ±√[(1 – cosθ)(1 + cosθ)]

Product-Sum Trigonometric Identities

The product-sum trigonometric identities change the sum or difference of sines or cosines into a product of sines and cosines.

Sin A + Sin B = 2 Sin(A+B)/2 . Cos(A-B)/2
Cos A + Cos B = 2 Cos(A+B)/2 . Cos(A-B)/2
Sin A – Sin B = 2 Cos(A+B)/2 . Sin(A-B)/2
Cos A – Cos B = -2 Sin(A+B)/2 . Sin(A-B)/2

Trigonometric Identities of Products

These identities are:

Sin A. Sin B = [Cos (A – B) – Cos (A + B)]/2
Sin A. Cos B = [Sin (A + B) – Sin (A – B)]/2
Cos A. Cos B = [Cos (A + B) – Cos (A – B)]/2

Trigonometric Identities Proofs

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

Trigonometric Identities

Applying Pythagoras Theorem for the given triangle, we have

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

AC² = AB²+BC²………………………..(1)

Let us prove the three Pythagoras trigonometric identities, which are commonly used.

Trigonometric Identity 1

Now, divide each term of equation (1) by AC², we have

$\begin{array}{l}\frac{AC^2}{AC^2} = \frac{AB^2}{AC^2}~+~\frac{BC^2}{AC^2}\end{array} $

$\begin{array}{l}\Rightarrow \frac{AB^2}{AC^2}+\frac{BC^2}{AC^2} = 1\end{array} $

$\begin{array}{l}\Rightarrow (\frac{AB}{AC})^2+(\frac{BC}{AC})^2 = 1 …(2)\end{array} $

We know that,

$\begin{array}{l}(\frac{AB}{AC})^2 = \cos a\ and\ (\frac{BC}{AC})^2 = \sin a \end{array} $

Thus equation (2) can be written as-

sin²a + cos²a = 1

Identity 1 is valid for angles 0 ≤ a ≤ 90.

Trigonometric Identity 2

Now Dividing the equation (1) by AB², we get

$\begin{array}{l}\frac{AC^2}{AB^2}= \frac{AB^2}{AB^2}+\frac{BC^2}{AB^2}\end{array} $

$\begin{array}{l}\Rightarrow \frac{AC^2}{AB^2}= 1+\frac{BC^2}{AB^2}\end{array} $

$\begin{array}{l}\Rightarrow (\frac{AC}{AB})^2 = 1+(\frac{BC}{AB})^2 …(3)\end{array} $

By referring to trigonometric ratios, it can be seen that:

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

Similarly,

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

Replacing the values of AC/AB and BC/AB in equation (3) gives,

1+tan² a = sec² a

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

Trigonometric Identity 3

Dividing the equation (1) by BC², we get

$\begin{array}{l}\frac{AC^2}{BC^2} = \frac{AB^2}{BC^2}~ +~\frac{BC^2}{BC^2}\end{array} $

$\begin{array}{l}\Rightarrow \frac{AC^2}{BC^2} = \frac{AB^2}{BC^2}+1\end{array} $

$\begin{array}{l}\Rightarrow (\frac{AC}{BC})^2 = (\frac{AB}{BC})^2+1 …(4)\end{array} $

By referring to trigonometric ratios, it can be seen that:

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

Also,

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

Replacing the values of AC/BC and AB/BC in the equation (4) gives,

cosec²a = 1 + cot²a

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 such that, 0° < a ≤ 90°.

Triangle Identities (Sine, Cosine, Tangent rule)

If the identities or equations are applicable for all the triangles and not just for right triangles, then they are the triangle identities. These identities will include:

If A, B and C are the vertices of a triangle and a, b and c are the respective sides, then;

According to the sine law or sine rule,

$\begin{array}{l}\frac{a}{Sin A} = \frac{b}{Sin B} = \frac{c}{Sin C}\end{array} $

$\begin{array}{l}\frac{Sin A}{a} = \frac{Sin B}{b} = \frac{Sin C}{c}\end{array} $

According to cosine law,

$\begin{array}{l}c^{2}=a^{2}+b^{2}-2 a b \cos C\end{array} $

$\begin{array}{l}cos~ C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}\end{array} $

According to tangent law,

$\begin{array}{l}\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}\end{array} $

Video Lessons on Trigonometric Identities

Trigonometric Identities Questions

Trigonometric Functions Questions

Solved Examples on Trigonometric Identities

Go through the below problem which is solved by using the trigonometric identities.

Example 1:

Consider a triangle ABC, right-angled at B. The length of the base, AB = 4 cm and length of perpendicular BC =3 cm. Find the value of sec A.

Solution:

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

tan A = 3/4

Now, using the trigonometric identity: 1+tan² a = sec² a

sec² A = 1 + (3/4)²

sec^{2 A = 25/16}

sec A = ±5/4

Since, the ratio of lengths is positive, we can neglect sec A = 5/4.

Therefore, sec A = 5/4

Example 2: (1 – sin A)/(1 + sin A) = (sec A – tan A)²

Solution: Let us take the Left hand side of the equation.

L.H.S = (1 – sin A)/(1 + sin A)

Multiply both numerator and denominator by (1 – sin A)

= (1 – sin A)²/(1 – sin A) (1 + sin A)

= (1 – sin A)²/(1 – sin² A)

= (1 – sin A)²/(cos² A), [Since sin² θ + cos² θ = 1 ⇒ cos² θ = 1 – sin² θ]

= {(1 – sin A)/cos A}²

= (1/cos A – sin A/cos A)²

= (sec A – tan A)²

= R.H.S.

Example 3: Prove that: 1/(cosec A – cot A) – 1/sin A = 1/sin A – 1/(cosec A + cot A)

Solution: 1/(cosec A – cot A) – 1/sin A = 1/sin A – 1/(cosec A + cot A)

Now rearrange the following, such that;

1/(cosec A – cot A) + 1/(cosec A + cot A) = 2/Sin A

Now let us take the L.H.S.

= 1/(cosec A – cot A) + 1/(cosec A + cot A)

= (cosec A + cot A + cosec A – cot A)/(cosec² A – cot² A)

= (2 cosec A)/1 [cosec² A = 1 + cot² A ⇒ cosec²A – cot² A = 1]

= 2/sin A [cosec A = 1/sin A]

Hence, proved.

Trigonometric Identities Practice Questions

Solve the below practice questions based on the trigonometry identities that will help in understanding and applying the formulas in an effective way.

Express the ratios cos A, tan A and sec A in terms of sin A.
Prove that sec A (1 – sin A)(sec A + tan A) = 1.
Find the value of 7 sec²A – 7 tan²A.
Show that (sin A + cosec A)² + (cos A + sec A)² = 7 + tan²A + cot²A

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

Frequently Asked Questions on Trigonometric Identities

What are three main functions in trigonometry?

The three main functions of trigonometry are Sine, Cosine and Tangent.
Sin θ = Opposite / Hypotenuse
Cos θ = Adjacent/Hypotenuse
Tan θ = Opposite/Adjacent

What are the basic 8 trigonometric identities?

The basic trigonometric identities are:
Cosec θ = 1/Sin θ
Sec θ = 1/Cos θ
Cot θ = 1/Tan θ
Tan θ = Sin θ/Cos θ
Cot θ = Cos θ/Sin θ
Sin2θ + Cos2 θ = 1
1 + tan² θ = sec² θ

What are the Pythagoras identities?

The three Pythagoras identities are:
sin² a + cos² a = 1
1+tan² a = sec² a
cosec² a = 1 + cot² a

What is the reciprocal of sine function?

The reciprocal of sine function is cosec function.

What is the value of sin 2A?

Sin 2 A = 2 Sin A Cos A
Sin 2 A = (2 tan A)/(1 + tan² A)

MATHS Related Links
Coordinate Geometry	Maths Syllabus
Greatest Integer Function	Inverse Tan
Vector Components	Law Of Sines
Value Of Log 10	Multiplication Sums For Class 4
Power Set	What Are Natural Numbers

MATHS Related Links

Coordinate Geometry

Maths Syllabus

Greatest Integer Function

Multiplication Sums For Class 4

Power Set

What Are Natural Numbers

3 Comments

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need some solved examples please help

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