Please Wait... While Loading Full Video#### class-10 chapter -8

### INTRODUCTION TO TRIGONOMETRY

`☼` 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.`

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

Class 10 Chapter 8 Example 12

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

Class 10 Chapter 8 Example 13

`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

Class 10 Chapter 8 Example 14

`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

Class 10 Chapter 8 Example 15

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.