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# Trigonometric Identities - Class 10

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Last updated date: 10th Aug 2024
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## What are Trigonometric Identities?

Trigonometry is a branch of mathematics that deals with the study of measuring the sides and angles of a triangle. The concept of trigonometry is completely based on right triangles. There are six trigonometric ratios, namely sine, cosine, tangent, cotangent, secant, and cosecant of a reference angle. All these trigonometric ratios are expressed as the ratios of the hypotenuse, base and perpendicular side of a right triangle.

Trigonometric identities - Class 10 establishes the relationship between these different trigonometric ratios. These trigonometric identities are the basic formulae that are true for all the values of the reference angle.

There are six trigonometric ratios namely sine, cosine, tangent cotangent, secant, and cosecant. The definition of each trigonometric ratio for a reference angle ‘θ’ is mentioned in terms of the sides of a right triangle. They are:

### Basic Trigonometric Ratios

 Property Mathematical value sin θ Perpendicular/Hypotenuse cos θ Base/Hypotenuse tan θ Perpendicular/Base sec θ Hypotenuse/Base cosec θ Hypotenuse/Perpendicular cot θ Base/Perpendicular

### Trigonometric Identities Class 10 List

Trigonometric identities can be used to solve trigonometric problems easily by reducing the number of computational steps.

There are basically three trigonometric identities that you study in Class 10. They are:

• $sin^{2} a + cos^{2} a = 1$

• $sec^{2}a = 1 + tan^{2}a$

• $cosec^{2}a = 1 + cot^{2}a$

Here, A is the reference angle taken for a right-angled triangle.

### Derivation of Trigonometric Identities Class 10

Let us consider a right-angled triangle ABC whose sides are AB, BC, and AC respectively. The triangle is right-angled at C. Considering ‘α’ as the reference angle, the opposite side is BC and the adjacent side is AC. AB is the longest side of the right triangle and is called the hypotenuse.

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Applying Pythagoras theorem to the above triangle,

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

On dividing the entire equation by the square of hypotenuse i.e.$AB^{2}$, we get,

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

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

By the definition of basic trigonometric functions,

$Sin \alpha = \frac{Opposite}{Hypotenuse} = \frac{BC}{AB} \, and \, Cos \alpha = \frac{Adjacent}{Hypotenuse} = \frac{AC}{AB}$

Substituting these values in the above equation,

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

The equation can be rewritten to give the first one among the trigonometric identities class 10 as:

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

This trigonometric identity is true for all angles ‘α’ such that 0° ≤ α ≤ 90°.

Again, using the same figure and equation 1, $AB^{2} = AC^{2} + BC^{2}$.

On dividing the equation by the square of the adjacent side i.e. ${AC^{2}}$, we get,

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

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

By the definition of basic trigonometric functions,

$tan \alpha = \frac{Opposite}{Adjacent} = \frac{BC}{AC} \, and \, Sec \alpha = \frac{Hypotenuse}{Adjacent} = \frac{AB}{AC}$

Substituting these values in the above equation,

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

The equation can be rewritten to give the second one among the trigonometric identities class 10 as,

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

This trigonometric identity is true for all angles ‘α’ such that 0° ≤ α < 90°.

Again, using the same figure and equation 1, $AB^{2} = AC^{2} + BC^{2}$

On dividing the equation by the square of the opposite side i.e. $BC^{2}$, we get,

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

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

By the definition of basic trigonometric functions,

$cot \alpha = \frac{Adjacent}{Opposite} = \frac{AC}{BC} \, and \, Cosec \alpha = \frac{Hypotenuse}{Opposite} = \frac{AB}{BC}$

Substituting these values in the above equation,

$(cosec \alpha )^{2} = (cot \alpha )^{2} + 1$

The equation can be rewritten to give the third one among the trigonometric identities class 10 as,

$cosec^{2} \alpha = 1 + cot^{2} \alpha$

This trigonometric identity is true for all angles ‘α’ such that 0° < α ≤ 90°.

### Trigonometric Identities Class 10 Problems

1. Find the value of 1 - Sin2 A.

Solution:

Using the identity, $sin^{2}A + cos^{2}A = 1$

Substituting the value of 1 in the given equation,

$1 - Sin^{2} A = (Sin^{2}A + Cos^{2}A) - Sin^{2}A$

$= Cos^{2} A$

2. Prove that $Sec^{2}P - tan^{2}P - Cosec^{2}P + Cot^{2}P = 0$.

Solution:

Using the identities,

• $sec^{2}a = 1 + tan^{2}a$

• $cosec^{2}a = 1 + cot^{2}a$

Substituting the value of sec^{2}A and cosec^{2}A in the given equation,

$Sec^{2}P - tan^{2}P - Cosec^{2}P + Cot^{2}P = (1 + tan^{2}P) - tan^{2}P - (1 + Cot^{2}P) + Cot^{2}P$

$= 1 + 0 - 1 - Cot^{2}P + Cot^{2}P$

= 0

### Conclusion

These trigonometry identities are the basis of trigonometry and can be used to express each trigonometric ratio in terms of other trigonometric ratios. Also, if you know the value of any trigonometry ratio, then you can find the values of other trigonometric ratios.

## FAQs on Trigonometric Identities - Class 10

1. What is the relation between different trigonometric ratios?

Here is the relation:

• tan A = sin A/ cos A = 1/ cot A

• cot A = cos A/ sin A

• sec A = 1/ cos A

• cosec A = 1/ sin A

2. How to prove the above discussed trigonometric identities -  Class 10?

For identity, $cos^{2}\theta + sin^{2}\theta = 1$

 If θ = 0 θ = 90 $cos^{2}0 + sin^{2}0 = 1$ $1^{2} + 0^{2} = 1$$1 + 0 = 1$$1 = 1$ $cos^{2}90 + sin^{2}90 = 1$$0^{2} + 1^{2} = 1$$0 + 1 = 1$$1 = 1$

For identity, $sec^{2}a = 1 + tan^{2}a$

 If θ = 0 θ = 90 $1 + tan^{2}0 = sec^{2}0$$1 + 0^{2} = 12$$1 = 1$ $1 + tan^{2}90 = sec^{2}90$$1 + \infty = \infty$$\infty = \infty$

For identity, $cosec^{2}a = 1 + cot^{2}a$

 If θ = 0 θ = 90 $1 + cot^{2}0 = cosec^{2}0$ $1 + \infty = \infty$$\infty = \infty$ $1 + cot^{2}90 = cosec^{2}90$ $1 + 0^{2} = 1^{2}$$1 = 1$