Trigonometric Identities

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

The six basic trigonometric ratios 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. Let us discuss the list of trigonometry identities, its derivation and problems solved using the important identities.

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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Trigonometric Identities List

There are various identities in trigonometry which are used for many trigonometric problems. Let us see all the fundamental trigonometric identities here.

Reciprocal Identities

• 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 Identities

• sin² a + cos² a = 1
• 1+tan² a  = sec² a
• cosec² a = 1 + cot² a

Ratio Identities

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

Opposite Angle Identities

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

Complementary Angles Identities

• 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

Angle Sum and Difference 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
• \(\tan (\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 – \tan \alpha. \tan \beta}\)
• \(\tan (\alpha – \beta) = \frac{\tan \alpha – \tan \beta}{1 + \tan \alpha. \tan \beta}\)

Trigonometric Identities Formula

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 the 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

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

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

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

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

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

sin² a + cos² a = 1

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

Trigonometric Identity 2

Now Dividing the equation (1) by AB², 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,

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

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

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

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

cosec² a = 1 + cot² a

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

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:

• Sine law
• Cosine law
• Tangent law

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, 

\(\frac{a}{Sin A} = \frac{b}{Sin B} = \frac{c}{Sin C}\)

Or

\(\frac{Sin A}{a} = \frac{Sin B}{b} = \frac{Sin C}{c}\)

According to cosine law,

\(c^{2}=a^{2}+b^{2}-2 a b \cos C\)

Or

\(cos~ C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}\)

According to tangent law,

\(\frac{a-b}{a+b}=\frac{\tan \left(\frac{A-B}{2}\right)}{\tan \left(\frac{A+B}{2}\right)}\)

Problems and Solutions

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 ⇒ cosec² A – cot² A = 1]

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

Hence, proved.

Trigonometric Identities Questions

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

1. Express the ratios cos A, tan A and sec A in terms of sin A.
2. Prove that sec A (1 – sin A)(sec A + tan A) = 1.
3. Find the value of 7 sec²A – 7 tan²A.
4. 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.