Innings2
Create New Account

Sign in to Innings2

or

New Account     Reset Password     

Chapter 11: Trigonometry

    Introduction

    Trigonometric ratios

    Trigonometric Ratios of Some Specific Angles

    Trigonometric Ratios Of Complementary Angles

    Trigonometric Identities

    Summary

    Enhanced Curriculum Support

    Easy Level Worksheet

    Moderate Level Worksheet

    Hard Level Worksheet

Powered by Innings 2

Reset Progress

This will delete your progress and chat data for all chapters in this course, and cannot be undone!

Glossary

Select one of the keywords on the left…

Chapter 11: Trigonometry > Trigonometric Identities

Trigonometric Identities

Trigonometric identities are important to learn because they allow us to solve equations involving trigonometric functions. These identities are based on the fundamental relationships between the angles and sides of triangles, and they can be used to simplify expressions, to find the exact values of trigonometric functions, and to prove other mathematical results. They are a fundamental tool in mathematics, and they are used in a wide range of applications, from geometry and calculus to engineering and physics.

Let us look at identifying some trigonometric identities.

Given the triangle ABC, the identity we already know is AC2 + BC2 = (1)

If we have to rephrase the above equation in terms of sin and cos, how can we do it? We know that sin and cos have ratios with the hypotenuse in the denominator. So, we need to divide each term in the equation with:

+ =

cosA2 + sinA2 =

cos2A + sin2A = (2)

This is true for all A such that 0° ≤ A ≤ 90°. So, this is a trigonometric identity.

If we have to get an identity in terms of tan we need to divide AC2 + BC2 = AB2 with the square of

+ tan2 A = sec2 A (3)

Is this equation true for A = 0° ?

, it is. What about A = 90°? Well, tan A and sec A are not defined for A = 90°. So, (3) is true for all A such that 0° ≤ A ≤ °.

Let us see what we get on dividing (1) by BC2. We get AB2BC2 = BC2BC2 + AC2BC2

i.e., ABBC2 = BCBC2 + ACBC2

i.e.,cot2A + = cosec2 A (4)

Note that cosec A and cot A are not defined for A = °.

Therefore (4) is true for all A such that 0° < A ≤ °.

Using these identities, we can express each trigonometric ratio in terms of other trigonometric ratios, i.e., if any one of the ratios is known, we can also determine the values of other trigonometric ratios.

Let us see how we can do this using these identities.

If tan A = 13, then, cot A = 3,3,13.

Look above and see which identity we can use. We need to choose an identity which has tan in it.

Since, sec2 A = 1 + tan2 A = 1 + 13 = , sec A = 23

The inverse of sec A is nothing but .

So cos A =32.

Now that we have cos A, we need to pick an identity which has cos in it.

sin2 A + cos2 A = .

So sinA = 1−cos2A,1−tan2A,1−cos2A.

Example 9

9. Express the ratios cos A, tan A and sec A in terms of sin A.

Solution : Since cos2A + sin2A =

Therefore, cos2A = –

cos A = ±1−sin2A

This gives cos A = 1−sin2A (as squareroot can't be )

Hence, tan A = sin Acos A = and sec A = 1(cos A) =

Example 10

10. Prove that sec A (1 – sin A)(sec A + tan A) = 1.

Solution :

LHS = secA1−sinAsecA+tanA = 1−sinA+sinAcosA

= 1−sinA1+sinAcos2A = = = = RHS

Example 11

11. Prove that cotA−cosAcotA+cosA = cosecA−1cosecA+1

Solution :

LHS = cotA−cosAcotA+cosA = cosAsinA−cosAcosAsinA+cosA

= cosA1sinA−1cosA1sinA+1 =

= = RHS

Example 12

12. Prove that sinθ−cosθ+1sinθ+cosθ−1 = 1secθ−tanθ , using the identity sec2θ=1+tan2θ.

Solution : Since we will apply the identity involving sec θ and tan θ, let us first convert the LHS (of the identity we need to prove) in terms of sec θ and tan θ by dividing numerator and denominator by θ.

LHS = sinθ−cosθ+1sinθ+cosθ−1 = tanθ+1−secθ

= tanθ+secθ−1tanθ−secθ+1 = tanθ+secθ−1tanθ−secθtanθ−secθ+1tanθ−secθ

= tan2θ−sec2θ−tanθ−secθtanθ−secθ+1tanθ−secθ

=

= = 1secθ−tanθ

which is the RHS of the identity, we are required to prove.

Trigonometric Tools

Some of the tools required to solve trignometric ratio problems

1)Reciprocal rule:

sinA =1cosecA

cosA =1secA

tanA = 1(cot A)

2)Quotient rule:

tanA =sinAcosA

cotA =(cos A)(sin A)

3)Identities:

sin2A+cos2A=1 => cos2A=1−sin2A and sin2A=1−cos2A

1+tan2A=sec2A⇒sec2A−tan2A=1 => tan2A=sec2A−1

cot2A+1=cosec2A⇒cosec2A−cot2A=1 => cot2A=cosec2A−1

4)Algebraic identities:

a+b2=a2+2ab+b2

a−b2=a2−2ab+b2

a2−b2=a+ba−b

a+b3=a3+b3+3aba+b

a−b3=a3−b3−3aba−b

a3−b3=a−ba2+ab+b2

Tips to solve problems using Trignometric ratios and identities:

  1. Always start from the complex side no. of terms,no. of operations,trignometric ratio,presence of square root in the question.
  • Eg: 1+cosA1−cosA=cosecA+cotA;

start from

as it has

  1. express in terms of sin & cos when different terms are seen.
  • Eg: tanA1−cotA+cotA1−tanA=1+sinAcosecA;
  1. Combine terms into single fraction.
  • Eg: cosA1+sinA+1+sinAcosA=2secA;
  1. Use Pythagoras theorem
  • Eg: sin2A1−cosA=1+secAsecA
  1. Solve L.H.S and R.H.S seperately and prove they are equal.
  • Eg: cosecA−sinA·secA−cosA=1tanA+cotA
  1. Expand / Factorize / Simplify / cancel common terms if needed using above Trigonometric Tools.