RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities - GMS - Learning Simply
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RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities

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RD Sharma Solutions for Class 10 Maths Chapter 6 Trigonometric Identities

The branch of mathematics which deals with the measurement of the sides and the angles of a triangle is trigonometry. We know that by now this topic would already seem difficult and complicated as its completely new to you. So, in order to make your learning process smooth and hassle-free the RD Sharma Solutions prepared by Goyanka Maths Study will help students get the correct understanding of various chapters in the book.

Trigonometric Identities is the 6th chapter of RD Sharma Class 10 which has two exercises and its solved answers with detailed explanations are given here RD Sharma Solutions for Class 10. The previous chapter was about trigonometric ratios and relations between them. But this chapter will be about proving some trigonometric identities and use them to prove other useful trigonometric identities.


Access the RD Sharma Solutions For Class 10 Maths Chapter 6 – Trigonometric Identities

RD Sharma Solutions for Class 10 Maths Chapter 6 Exercise 6.1 Page No: 6.43

Prove the following trigonometric identities:

1. (1 – cos2 A) cosec2 A = 1

Solution:

Taking the L.H.S,

(1 – cos2 A) cosec2 A

= (sin2 A) cosec2 A [∵ sin2 A + cosA = 1 ⇒1 – sin2 A = cos2 A]

= 12

= 1 = R.H.S

– Hence Proved

2. (1 + cot2 A) sin2 A = 1 

Solution: 

By using the identity,

cosecA – cot2 A = 1 ⇒ cosecA = cot2 A + 1

Taking,

L.H.S = (1 + cot2 A) sin2 A

= cosec2 A sin2 A

= (cosec A sin A)2

= ((1/sin A) × sin A)2

= (1)2

= 1

= R.H.S

– Hence Proved

3. tanθ cosθ = 1 − cosθ 

Solution: 

We know that,

sinθ + cosθ = 1

Taking,

L.H.S = tanθ cosθ

= (tan θ × cos θ)2

= (sin θ)2

= sinθ

= 1 – cosθ

= R.H.S

– Hence Proved

4. cosec θ √(1 – cos2 θ) = 1

Solution:

Using identity,

sinθ + cosθ = 1  ⇒ sinθ = 1 – cosθ

Taking L.H.S,

L.H.S = cosec θ √(1 – cos2 θ)

= cosec θ √( sinθ)

= cosec θ x sin θ

= 1

= R.H.S

– Hence Proved

5. (secθ − 1)(cosecθ − 1) = 1 

Solution:

Using identities,

(secθ − tanθ) = 1 and (cosecθ − cotθ) = 1

We have,

L.H.S = (secθ – 1)(cosec2θ – 1)

= tan2θ × cot2θ

= (tan θ × cot θ)2

= (tan θ × 1/tan θ)2

= 12

= 1

= R.H.S

– Hence Proved

6. tan θ + 1/ tan θ = sec θ cosec θ

Solution:

We have,

L.H.S = tan θ + 1/ tan θ

= (tan2 θ + 1)/ tan θ

= sec2 θ / tan θ [∵ secθ − tanθ = 1]

= (1/cos2 θ) x 1/ (sin θ/cos θ) [∵ tan θ = sin θ / cos θ]

= cos θ/ (sin θ x cos2 θ)

= 1/ cos θ x 1/ sin θ

= sec θ x cosec θ

= sec θ cosec θ

= R.H.S

– Hence Proved

7. cos θ/ (1 – sin θ) = (1 + sin θ)/ cos θ

Solution:

We know that,

sinθ + cosθ = 1

So, by multiplying both the numerator and the denominator by (1+ sin θ), we get

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 1

L.H.S =

= R.H.S

– Hence Proved

8. cos θ/ (1 + sin θ) = (1 – sin θ)/ cos θ

Solution:

We know that,

sinθ + cosθ = 1

So, by multiplying both the numerator and the denominator by (1- sin θ), we get

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 2

L.H.S =

= R.H.S

– Hence Proved

9. cosθ + 1/(1 + cotθ) = 1

Solution:

We already know that,

cosecθ − cotθ = 1 and sinθ + cosθ = 1

Taking L.H.S,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 3

= cos2 A + sin2 A

= 1

= R.H.S

– Hence Proved

10. sinA + 1/(1 + tan A) = 1

Solution:

We already know that,

secθ − tanθ = 1 and sinθ + cosθ = 1

Taking L.H.S,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 4

= sin2 A + cos2 A

= 1

= R.H.S

– Hence Proved

11.
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 5

Solution:

We know that, sinθ + cosθ = 1

Taking the L.H.S,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 6

= cosec θ – cot θ

= R.H.S

– Hence Proved

12. 1 – cos θ/ sin θ = sin θ/ 1 + cos θ

Solution:

We know that,

sinθ + cosθ = 1

So, by multiplying both the numerator and the denominator by (1+ cos θ), we get

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 7

= R.H.S

– Hence Proved

13. sin θ/ (1 – cos θ) = cosec θ + cot θ

Solution:

 

Taking L.H.S,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 8

= cosec θ + cot θ

= R.H.S

– Hence Proved

14. (1 – sin θ) / (1 + sin θ) = (sec θ – tan θ)2

Solution:

Taking the L.H.S,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 9

= (sec θ – tan θ)2

= R.H.S

– Hence Proved

15. R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 10

Solution:

Taking L.H.S,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 11

= cot θ

= R.H.S

– Hence Proved

16. tanθ − sinθ = tanθ sinθ 

Solution:

Taking L.H.S,

L.H.S = tanθ − sinθ 

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 12

= tanθ sinθ

= R.H.S

– Hence Proved

17. (cosec θ + sin θ)(cosec θ – sin θ) = cot2θ + cos2θ 

Solution:

Taking L.H.S = (cosec θ + sin θ)(cosec θ – sin θ)

On multiplying we get,

= cosec2 θ – sin2 θ

= (1 + cot2 θ) – (1 – cos2 θ) [Using cosecθ − cotθ = 1 and sinθ + cosθ = 1]

= 1 + cot2 θ – 1 + cos2 θ

= cot2 θ + cos2 θ

= R.H.S

– Hence Proved

18. (sec θ + cos θ) (sec θ – cos θ) = tanθ + sinθ 

Solution:

Taking L.H.S = (sec θ + cos θ)(sec θ – cos θ)

On multiplying we get,

= sec2 θ – sin2 θ

= (1 + tan2 θ) – (1 – sin2 θ) [Using secθ − tanθ = 1 and sinθ + cosθ = 1]

= 1 + tan2 θ – 1 + sin2 θ

= tan θ + sin 2 θ

= R.H.S

– Hence Proved

19. sec A(1- sin A) (sec A + tan A) = 1

Solution:

Taking L.H.S = sec A(1 – sin A)(sec A + tan A)

Substituting sec A = 1/cos A and tan A =sin A/cos A in the above we have,

L.H.S = 1/cos A (1 – sin A)(1/cos A + sin A/cos A)

= 1 – sin2 A / cos2 A [After taking L.C.M]

= cos2 A / cos2 A [∵ 1 – sin2 A = cos2 A]

= 1

= R.H.S

– Hence Proved

20. (cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1 

Solution:

Taking L.H.S = (cosec A – sin A)(sec A – cos A)(tan A + cot A)

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 13

= (cos2 A/ sin A) (sin2 A/ cos A) (1/ sin A cos A) [∵ sinθ + cosθ = 1]

= (sin A  cos A)  (1/ cos A sin A)

= 1

= R.H.S

– Hence Proved

21. (1 + tanθ)(1 – sin θ)(1 + sin θ) = 1

Solution:

Taking L.H.S = (1 + tan2θ)(1 – sin θ)(1 + sin θ)

And, we know sin2 θ + cos2 θ = 1 and sec2 θ – tan2 θ = 1

So,

L.H.S = (1 + tan2 θ)(1 – sin θ)(1 + sin θ)

= (1 + tanθ){(1 – sin θ)(1 + sin θ)}

= (1 + tanθ)(1 – sinθ)

= secθ (cos2 θ)

= (1/ cos2 θ) x cos2 θ

= 1

= R.H.S

– Hence Proved

22. sinA cotA + cosA tanA = 1

Solution:

We know that,

cotA = cosA/ sin2 A and tan2 A = sin2 A/cos2 A

Substituting the above in L.H.S, we get

L.H.S = sinA cotA + cosA tanA

= {sinA (cosA/ sin2 A)} + {cosA (sin2 A/cos2 A)}

= cosA + sin2 A

= 1 [∵ sinθ + cosθ = 1]

= R.H.S

– Hence Proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 14

23.

Solution:

(i) Taking the L.H.S and using sinθ + cosθ = 1, we have

L.H.S = cot θ – tan θ

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 15

= R.H.S

– Hence Proved

(ii) Taking the L.H.S and using sinθ + cosθ = 1, we have

L.H.S = tan θ – cot θ

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 16

= R.H.S

– Hence Proved

24. (cos2 θ/ sin θ) – cosec θ + sin θ = 0

Solution:

Taking L.H.S and using sinθ + cosθ = 1, we have

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 17

= – sin θ + sin θ

= 0

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 1825.

Solution:

Taking L.H.S,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 19

= 2 sec2 A

= R.H.S

  • Hence proved

26. R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 20

Solution:

Taking the LHS and using sinθ + cosθ = 1, we have

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 21

= 2/ cos θ

= 2 sec θ

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 22

27.

Solution:

 

Taking the LHS and using sinθ + cosθ = 1, we have

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 23

= R.H.S

  • Hence proved

28.R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 24

Solution:

Taking L.H.S,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 25

Using secθ − tanθ = 1 and cosecθ − cotθ = 1

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 26

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 32

29.

Solution:

Taking L.H.S and using sinθ + cosθ = 1, we have

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 34
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 33

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 35

30.

Solution:

Taking LHS, we have

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 36

= 1 + tan θ + cot θ

= R.H.S

  • Hence proved

31. secθ = tanθ + 3 tanθ secθ + 1

Solution: 

From trig. Identities we have,

secθ − tanθ = 1

On cubing both sides,

(sec2θ − tan2θ)= 1

secθ − tanθ − 3secθ tanθ(secθ − tanθ) = 1

[Since, (a – b)3 = a3 – b3 – 3ab(a – b)]

secθ − tanθ − 3secθ tanθ = 1

⇒ secθ = tanθ + 3secθ tanθ + 1

Hence, L.H.S = R.H.S

  • Hence proved

32. cosecθ = cotθ + 3cotθ cosecθ + 1

Solution:

From trig. Identities we have,

cosecθ − cotθ = 1

On cubing both sides,

(cosecθ − cotθ)3 = 1

cosecθ − cotθ − 3cosecθ cotθ (cosecθ − cotθ) = 1

[Since, (a – b)3 = a3 – b3 – 3ab(a – b)]

cosecθ − cotθ − 3cosecθ cotθ = 1

⇒ cosecθ = cotθ + 3 cosecθ cotθ + 1

Hence, L.H.S = R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 3733.

Solution:

Taking L.H.S and using secθ − tanθ = 1 ⇒ 1 + tanθ = secθ

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 38

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 3934.

Solution:

Taking L.H.S and using the identity sin2A + cos2A = 1, we get

sin2A = 1 − cos2A

⇒ sin2A = (1 – cos A)(1 + cos A)

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 40

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 4135.

Solution:

We have,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 42

Rationalizing the denominator and numerator with (sec A + tan A) and using secθ − tanθ = 1 we get,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 43

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 44

36.

Solution:

We have,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 45

On multiplying numerator and denominator by (1 – cos A), we get

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 45

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 47

37. (i)

Solution:

Taking L.H.S and rationalizing the numerator and denominator with √(1 + sin A), we get

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 48

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 49

(ii)

Solution:

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 50

= 2 cosec A

= R.H.S

  • Hence proved

38. Prove that:

(i)R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 51

Solution:

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 52

= 2 cosec θ

= R.H.S

  • Hence proved

(ii)R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 53

Solution:

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 54

= R.H.S

  • Hence proved

(iii)R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 55

Solution:

Taking L.H.S and rationalizing the numerator and denominator with its respective conjugates, we get

 

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 56

= 2 cosec θ

= R.H.S

  • Hence proved

(iv)R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 57

Solution:

Taking L.H.S, we have

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 58

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 59

39.

Solution:

 

Taking LHS = (sec A – tan A)2 , we have

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 60

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 61

40.

Solution:

Taking L.H.S and rationalizing the numerator and denominator with (1 – cos A), we get

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 62

= (cosec A – cot A)2

= (cot A – cosec A)2

= R.H.S

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 63

41.

Solution:

Considering L.H.S and taking L.C.M and on simplifying we have,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 64

= 2 cosec A cot A = RHS

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 65

42.

Solution:

Taking LHS, we have

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 66

= cos A + sin A

= RHS

  • Hence proved

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 6743.

Solution:

Considering L.H.S and taking L.C.M and on simplifying we have,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.1 - 68

= 2 secA

= RHS

  • Hence proved

RD Sharma Solutions for Class 10 Maths Chapter 6 Exercise 6.2 Page No: 6.54

1. If cos θ = 4/5, find all other trigonometric ratios of angle θ. 

Solution:

 

We have,

cos θ = 4/5

And we know that,

sin θ = √(1 – cosθ)

⇒ sin θ = √(1 – (4/5)2)

= √(1 – (16/25))

= √[(25 – 16)/25]

= √(9/25)

= 3/5

∴ sin θ = 3/5

Since, cosec θ = 1/ sin θ

= 1/ (3/5)

⇒ cosec θ = 5/3

And, sec θ = 1/ cos θ

= 1/ (4/5)

⇒ cosec θ = 5/4

Now,

tan θ = sin θ/ cos θ

= (3/5)/ (4/5)

⇒ tan θ = 3/4

And, cot θ = 1/ tan θ

= 1/ (3/4)

⇒ cot θ = 4/3

2. If sin θ = 1/√2, find all other trigonometric ratios of angle θ.

Solution:

 

We have,

sin θ = 1/√2

And we know that,

cos θ = √(1 – sinθ)

⇒ cos θ = √(1 – (1/√2)2)

= √(1 – (1/2))

= √[(2 – 1)/2]

= √(1/2)

= 1/√2

∴ cos θ = 1/√2

Since, cosec θ = 1/ sin θ

= 1/ (1/√2)

⇒ cosec θ = √2

And, sec θ = 1/ cos θ

= 1/ (1/√2)

⇒ sec θ = √2

Now,

tan θ = sin θ/ cos θ

= (1/√2)/ (1/√2)

⇒ tan θ = 1

And, cot θ = 1/ tan θ

= 1/ (1)

⇒ cot θ = 1

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 1

3.

Solution:

Given,

tan θ = 1/√2

By using secθ − tanθ = 1,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 2
R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 3

4.

Solution:

Given,

tan θ = 3/4

By using secθ − tanθ = 1,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 4

sec θ = 5/4

Since, sec θ = 1/ cos θ

⇒ cos θ = 1/ sec θ

= 1/ (5/4)

= 4/5

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 5

So,

 

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 6

5.

Solution:

Given, tan θ = 12/5

Since, cot θ = 1/ tan θ = 1/ (12/5) = 5/12

Now, by using cosecθ − cotθ = 1

cosec θ = √(1 + cotθ)

= √(1 + (5/12))

= √(1 + 25/144)

= √(169/ 25)

⇒ cosec θ = 13/5

Now, we know that

sin θ = 1/ cosec θ

= 1/ (13/5)

⇒ sin θ = 5/13

Putting value of sin θ in the expression we have,

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 7

= 25/ 1

= 25

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 8

6.

Solution:

Given,

cot θ = 1/√3

Using cosecθ − cotθ = 1, we can find cosec θ

cosec θ = √(1 + cot2 θ)

= √(1 + (1/√3)2)

= √(1 + (1/3)) = √((3 + 1)/3)

= √(4/3)

⇒ cosec θ = 2/√3

So, sin θ = 1/ cosec θ = 1/ (2/√3)

⇒ sin θ = √3/2

And, we know that

cos θ = √(1 – sin2 θ)

= √(1 – (√3/2)2)

= √(1 – (3/4))

√((4 – 3)/4)

= √(1/4)

⇒ cos θ = 1/2

Now, using cos θ and sin θ in the expression, we have

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 9

= 3/5

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 10

7.

Solution:

Given,

cosec A = √2

Using cosecA − cotA = 1, we find cot A

R D Sharma Solutions For Class 10 Maths Chapter 6 Trigonometric Identities ex 6.2 - 11

= 4/2

= 2

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