Class 10 INTRODUCTION TO TRIGONOMETRY

### Topic to be covered

☼ Trigonometric Identities .

### Trigonometric Identities

You may recall that an equation is called an identity when it is true for all values of the variables involved. Similarly, an equation involving trigonometric ratios of an angle is called a "trigonometric identity," if it is true for all values of the angle(s) involved.

In this section, we will prove one trigonometric identity, and use it further to prove other useful trigonometric identities.

In Delta ABC, right-angled at B (see Fig. 8.22), we have :

AB^2 + BC^2 = AC^2 .................. (1)

Dividing each term of (1) by AC^2, we get

(AB^2)/(AC^2) + (BC^2)/(AC^2) = (AC^2)/(AC^2)

((AB)/(AC))^2 + ((BC)/(AC))^2 = ((AC)/(AC))^2

i.e., (cos A)^2 + (sin A)^2 = 1

i.e., cos^2 A + sin^2 A = 1 ......................... (2)

This is true for all A such that 0° ≤ A ≤ 90°. So, this is a trigonometric identity.
Let us now divide (1) by AB^2. We get

(AB^2)/(AB^2) + (BC^2)/(AB^2) = (AC^2)/(AB^2)

((AB)/(AB))^2+((BC)/(AB))^2 = ((AC)/(AB))^2

i.e., 1 + tan^2 A = sec^2 A .....................(3)

Is this equation true for A = 0°? Yes, 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 < 90°.

Let us see what we get on dividing (1) by BC^2. We get

(AB^2)/(BC^2) + (BC^2)/(BC^2) = (AC^2)/(BC^2)

((AB)/(BC))^2+((BC)/(BC))^2 = ((AC)/(BC))^2

i.e., cot^2 A + 1 = cosec^2 A ...................... (4)

"Note" that cosec A and cot A are not defined for A = 0°. Therefore (4) is true for all A such that 0° < A ≤ 90°.

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. Suppose we know that

tanA = 1/sqrt3 , Then, cot A = sqrt3

Since, sec^2 A = 1 + tan^2 A = 1+1/3 = 4/3 , secA = 2/sqrt3  and cosA = sqrt3/2

Again, sin A = sqrt(1-cos^2 A) = sqrt(1-3/4) = 1/2 Therefore, cosec A = 2.
Q 3230601512

Express the ratios cos A, tan A and sec A in terms of sin A.
Class 10 Chapter 8 Example 12
Solution:

Since cos^2 A + sin^2 A = 1, therefore,
cos^2 A = 1 – sin^2 A, i.e., cos A = pm sqrt(1-sin^2 A)

This gives cosA = sqrt(1-sin^2A) (Why?)

Hence tanA = (sinA)/(cosA) = (sinA)/sqrt(1-sin^2A) and secA = 1/(cosA) = 1/sqrt(1-sin^2A)
Q 3250601514

Prove that sec A (1 – sin A)(sec A + tan A) = 1.
Class 10 Chapter 8 Example 13
Solution:

LHS = sec A (1 – sin A)(sec A + tan A) = (1/(cosA)) (1-sinA) (1/(cosA) + (sinA)/(cosA))

 = ((1-sinA)(1+sinA))/(cos^2A) = (1-sin^2A)/(cos^2A)

 = (cos^2 A)/(cos^2 A) = 1 = RHS
Q 3260601515

Prove that (cotA - cosA)/(cotA+cosA) = (cosecA-1)/(cosecA+1)
Class 10 Chapter 8 Example 14
Solution:

LHS = (cotA-cosA)/(cotA+cosA) = {(cosA)/(sinA) -cosA}/{(cosA)/(sinA)+cosA}

 = { cosA (1/(sinA)-1)}/{cosA(1/(sinA)+1)} = (1/(sinA)-1)/(1/(sinA)+1) = (cosecA-1)/(cosecA+1) = RHS
Q 3280601517

Prove that (sintheta-costheta+1)/(sintheta+costheta-1) = 1/(sectheta-tantheta) using the identity sec^2 theta = 1+tan^2 theta

Class 10 Chapter 8 Example 15
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 cos θ.

LHS = ( sintheta - costheta+1)/(sintheta+costheta-1) = (tantheta-1+sectheta)/(tantheta+1-sectheta)

= ((tantheta+sectheta)-1)/((tantheta-sectheta)+1) = {{(tantheta+sectheta)-1}(tantheta-sectheta)}/{{(tantheta-sectheta)+1}(tantheta-sectheta)}

= {(tan^2 theta-sec^2theta)-(tantheta-sectheta)}/{(tantheta-sectheta+1)(tantheta-sectheta)}

 = (-1-tantheta+sectheta)/{(tantheta-sectheta)(tantheta-sectheta)}

 = (-1)/(tantheta-sectheta) = 1/(sectheta-tantheta)

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