Trigonometry Class 10th Mathematics Gujarat Board Solution

Class 10^th Mathematics Gujarat Board Solution

Exercise 9.1

1. In ΔABC, m∠A = 90. If AB = 5, AC = 12 and BC = 13, find sinC, cosC, tanB, cosB,…
2. In ΔABC, m∠B = 90. If BC = 3 and AC = 5, find all the six trigonometric ratios…
3. If cosA = 4/5 find sinA and tanA.
4. If cosec θ = 13/5 find tan θ and cos θ.
5. If cosB = 1/3 , find the other five trigonometric ratios.
6. In ΔABC, m∠A = 90 and if AB : BC = 1 : 2 find sinB, cosC, tanC.
7. If tan θ = 4/3 , find the value of 5sintegrate heta +2costheta /3sintegrate heta…
8. If sec θ = 13/5 find the value of 5sintegrate heta +3costheta /5costheta…
9. If sinB = 1/2 prove that 3cosB - 4cos^3 B = 0.
10. If tanA = √3, verify that (1) sin^2 A + cos^2 A = 1 (2) sec^2 A - tan^2 A = 1…
11. If cos θ = 2 root 2/3 , verify that tan^2 θ - sin^2 θ = tan^2 θ⋅ sin^2 θ…
12. In ΔABC, m∠B = 90, AC + BC = 25 and AB = 5, determine the value of sinA, cosA…
13. In ΔABC, m∠C = 90 and m∠A = m∠B, (1) Is cosA = cosB? (2) Is tanA = tanB? (3)…
14. If 3cotA = 4, examine whether 1-tan^2a/1+tan^2a cos^2 A - sin^2 A.…
15. If pcot θ = q, examine whether psi ntheta -qcostheta /psi ntheta +qcostheta =…
16. State whether the following are true or false. Justify your answer: (1) sin θ =…

Exercise 9.2

1. cos60 = 1 - 2sin^2 30 = 2cos^2 30 - 1 = cos^2 30 - sin^2 30 Verify:…
2. sin60 = 2sin30 cos30 Verify:
3. sin60 = 2tan30/1+tan^230 Verify:
4. cos60 = 1-tan^230/1+tan^230 Verify:
5. cos90 = 4cos^3 30 - 3cos30 Verify:
6. sin30+tan45-cosec60/sec30+cos60+cot45 Evaluate:
7. 5cos^260+4sec^230-tan^245/sin^230+cos^230 Evaluate:
8. 2sin^2 30 cot30 - 3cos^2 60 sec^2 30 Evaluate:
9. 3cos^2 30 + sec^2 30 + 2cos0 + 3sin90 - tan^2 60 Evaluate:
10. In ΔABC, m∠B = 90, find the measure of the parts of the triangle other than the…
11. In a rectangle ABCD, AB = 20, m∠BAC = 60, calculate the length of side bar bc…
12. If θ is measure of an acute angle and cosθ = sinθ, find the value of 2tan^2 θ +…
13. If α is measure of acute angle and 3sinα = 2cosα, prove that (1-tan^2alpha…
14. sin(A + B) = sinA cosB + cosA sinB, If A = 30 and B = 60, verify that…
15. cos(A + B) = cosA cosB - sinA sinB If A = 30 and B = 60, verify that…
16. If sin(A - B) = sinA cosB - cosA sinB and cos(A - B) = cosA cosB + sinA sinB,…
17. State whether the following are true or false. Justify your answer: (1) The…

Exercise 9.3

1. cos18/sin72 Evaluate:
2. tan48 — cot42 Evaluate:
3. cosec32 — sec58 Evaluate:
4. cos70/sin20 + cos59 ⋅ cosec31 Evaluate:
5. sec70 sin20 — cos20 cosec70 Evaluate:
6. cos(40— theta) — sin(50 + theta) + cos^240+cos^250/sin^240+sin^250 Evaluate:…
7. cos70/sin20 + cos55cosec35/tan5tan25tan45tan65tan85 Evaluate:
8. cot12 ⋅ cot38 ⋅ cot52 ⋅ cot60 ⋅ cot78 Evaluate:
9. sin18/cos72 + √3 (tan10 tan30 tcm40 tan50 tan80- Evaluate:
10. cos70/sin20 + sin22/cos68 - cos38cosec52/tan18tan35tan60tan72tan55 Evaluate:…
11. sin48 sec42 + cos48 cosec42 = 2 Prove the following:
12. sin70/cos20 + cos6c^20/sec70 — 2cos70 cosec20 = 0 Prove the following:…
13. Prove the following:
14. cos (90-a) sin (90-a)/tan (90 - phi) = sin^2a Prove the following:…
15. Express the following in terms of trigonometric ratios of angles having measure…
16. For ΔABC, prove that (1) tan (a+c/2) = cot b/2 , (2) cos (b+c/2) = sin (a/2)…
17. If A + B = 90, prove that root tanatanb+tanacotb/sinasecb = seca
18. If 3 θ is the measure of an acute angle and sin30 = cos(θ — 26), then find the…
19. If 0 θ 90, θ, sinθ = cos30, then obtain the value of 2tan^2 θ — 1.…
20. If tanA = cotB, prove that A + B = 90, where A and B are measures of acute…
21. If sec2A = cosec(A — 42), where 2A is the measure of an acute angle, find the…
22. If 0 θ 90 and secθ = cosec60, find the value of 2cos^2 θ — 1.

Exercise 9

1. cos^2theta + 1/1+cot^2theta = 1 Prove the following by using trigonometric…
2. 2sin^2 θ + 4sec^2 θ + 5cot^2 θ + 2cos^2 θ — 4tan^2 θ — 5cosec^2 θ = 1 Prove the…
3. 1/1+costheta + 1/1-costheta = 2costheta c^2theta Prove the following by using…
4. Prove the following by using trigonometric identities:
5. root 1-sintegrate heta /1+sintegrate heta = sectheta -tantheta Prove the…
6. sectheta +tantheta /costheta c theta +cottheta = costheta c theta -cottheta…
7. cottheta +cosectheta -1/cottheta -cosectheta +1 = cosectheta +cottheta Prove the…
8. (sinθ + cosecθ)^2 + (cosθ + secθ)^2 = 7 + tan^2 θ + cot^2 θ. Prove the following…
9. 2sec^2 θ — sec^4 θ — 2cosec^2 θ + cosec^4 θ = cot^4 θ — tan^4 θ. Prove the…
10. (sinθ — secθ)^2 + (cosθ — cosecθ)^2 = (1 — secθ ⋅ cosecθ)^2 . Prove the…
11. Prove the following by using trigonometric identities:
12. tantheta -cottheta /sintegrate heta costheta = sec^2theta -costheta c^2theta =…
13. sectheta -tantheta /sectheta +tantheta = 1-2sectheta tantheta +tan^2theta Prove…
14. root sec^2theta +costheta c^3theta = tantheta +cottheta Prove the following by…
15. Prove the following by using trigonometric identities:
16. tantheta /1-cottheta + cottheta /1-tantheta = 1+tantheta +cottheta = 1+sectheta…
17. sin^4 θ - cos^4 θ = sin^2 θ - cos^2 θ = 2sin^2 θ - 1 = 1 - 2 cos^2 θ. Prove the…
18. tan^2 A — tan^2 B = Prove the following by using trigonometric identities:…
19. 2(sin^6 θ + cos^6 θ) — 3(sin^4 θ + cos^4 θ) + 1 = 0 Prove the following by…
20. If sinθ + cosθ = p and secθ + cosecθ = q, show that q(p^2 — 1) = 2p.…
21. If tanθ + sin = a and tanθ — sinθ = b, then prove that a^2 — b^2 = 4 root ab…
22. acosθ + bsinθ = p and asinθ— bcosθ = q, then prove that a^2 + b^2 = p^2 + q^2 .…
23. secθ + tanθ = p, then obtain the values of secθ, tanθ and sinθ in terms of p.…
24. sec38/cosec52 + 2/root 3 ⋅ tan17 tan38 tan60 tan52 tan73 — 3(sin^2 32 + sin^2…
25. - cottan (90 - theta) + cos6ctheta sectheta (90 - theta)
26. If sinA + cosA = √2 sin(90—A), then obtain the value of cotA.
27. If cosecθ = √2, then find the value of 2sin^2theta +3cot^2theta /4tan^2theta…
28. If tantheta = 8/15 then evaluate (1+sintegrate heta) (2-2sintegrate…
29. If costheta = b/root a^2 + b^2 , 0 θ 90, find the value of sinθ and tanθ.…
30. If θ is the measure of an acute angle such that bsinθ = acosθ, then…
31. Which of the following is correct for some 0 such that 0 ≤ θ 90?A. 1/sectheta…
32. If tantheta = 1/root 5 , then cosec^2theta -sec^2theta /cosec^2theta…
33. If tan^2theta = 8/7 , then the value of (1+sintegrate heta) (1-sintegrate…
34. If cottheta = 4/3 , then the value of costheta -sintegrate heta /costheta…
35. If cosecA = 4/3 and A + B = 90, then secB is ………..A. 3/4 B. 1/3 C. 4/3 D. 7/3…
36. If θ is the measure of an acute angle and √3 sinθ = cosθ, then θ is ____A. 30…
37. If tana = 5/12 then the value of (sinA + cosA) secA is …..A. 12/5 B. 7/12 C.…
38. If tantheta = 4/3 then the value of root 1-sintegrate heta /1+sintegrate heta…
39. In ΔABC, if m∠ABC = 90, m∠ACB = 45 and AC = 6, then area of ΔABC is …..A. 18…
40. If cos^2 45 - cos^2 30 = x ⋅ cos45 ⋅ sin45, then x is …….A. 2 B. 3/2 C. - 1/2…
41. If A and B are complementary angles, then sinA ⋅ secB is ____A. 1 B. 0 C. —1…
42. The value of tan20 tan25 tan45 tan65 tan70 is ………….A. —1 B. 1 C. 0 D. √3…
43. If 7θ and 2θ are measure of acute angles such that sin7θ = cos2θ then 2sin3θ —…
44. If A + B = 90, then cotacotb+cotatanb/sina secb - sin^2b/cos^2b is ….A. cot^2…
45. For ΔABC, sin (b+c/2) = ……..A. sin a/2 B. sinA C. cos a/2 D. cosA…
46. sin^4theta -cos^4theta /sin^2theta -cos^2theta = ………..A. 1 B. 2 C. 3 D. 0…
47. If 7cos^2 θ + 3sin^2 θ = 4, then cotθ is ___A. 7 B. 7/3 C. root 3 D. 1/root 3…
48. If tan5θ ⋅ tan4θ = 1, θ is ____A. 7 B. 3 C. 10 D. 9
49. If A and B are measures of acute angles and tanA = 1/root 3 and sinB = 1/2 ,…

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Exercise 9.1

Question 1.

In ΔABC, m∠A = 90. If AB = 5, AC = 12 and BC = 13, find sinC, cosC, tanB, cosB, sinB.

Answer:

we know that

Question 2.

In ΔABC, m∠B = 90. If BC = 3 and AC = 5, find all the six trigonometric ratios of ∠A.

Answer:

Pythagoras theorem

3² + AB² = 5²

⇒ AB = 4

we know that

Question 3.

If cosA =  find sinA and tanA.

Answer: 

Let AB = 4 and AC = 5

BC = 3 (by Pythagoras theorem)

we know that

⇒ 

⇒ 

Question 4.

If cosec θ = find tan θ and cos θ.

Answer:

Let ∠A be θ

BC = 5k, AC = 13k

⇒ AB = 12k (by Pythagoras theorem)

we know that

⇒ 

⇒ 

Question 5.

If cosB = , find the other five trigonometric ratios.

Answer: 

Let BC = k and AB = 3k

⇒ AC = (Pythagoras theorem)

we know that

Question 6.

In ΔABC, m∠A = 90 and if AB : BC = 1 : 2 find sinB, cosC, tanC.

Answer:

AB = k and BC = 2k

AC = (by Pythagoras theorem)

Question 7.

If tan θ = , find the value of 

Answer:

Let’s divide both numerator and denominator by cosθ, we get

put tanθ = in this equation.

Question 8.

If sec θ =  find the value of 

Answer:

Let ∠B = θ

⇒ AB = 5k, BC = 13k

⇒ AC = 12k (by Pythagoras theorem)

we know that

⇒ 

⇒ 

putting the above values in the given equations.

=

Question 9.

If sinB =  prove that 3cosB – 4cos³B = 0.

Answer:

Let AC = 1 and AB = 2.

⇒ 

∴ BC = √3 (by Pythagoras theorem) we know that

⇒ 

⇒ 

=

= 0

which is equal to the R.H.S

Question 10.

If tanA = √3, verify that

(1) sin²A + cos²A = 1

(2) sec²A – tan²A = 1

(3) 1 + cot²A = cosec²A

Answer:

Let BC = √3 and AC = 1

⇒ AB = 2 (by Pythagoras theorem)

we know that

⇒ 

⇒ 

⇒ 

⇒ 

⇒ 

(1) sin²A + cos²A = 1

⇒

(2) sec²A – tan²A = 1

⇒

(3) 1 + cot²A = cosec²A

⇒ L.H.S

⇒ 

R.H.S

⇒ 

L.H.S = R.H.S

Question 11.

If cos θ =  , verify that tan²θ – sin²θ = tan²θ⋅ sin²θ

Answer:

Let ∠B = θ

and BC = 2√2 and AB = 3

⇒ AC = 1 (by Pythagoras theorem)

we know that

⇒ tanθ =

⇒ sinθ =

L.H.S

tan² – sin²

by the above values of tanθ and sinθ

tan² – sin²

R.H.S

by the above values of tanθ and sinθ

tan²× sin²

Question 12.

In ΔABC, m∠B = 90, AC + BC = 25 and AB = 5, determine the value of sinA, cosA and tanA.

Answer:

AC = 25 – BC

AC² = AB² + BC² (by Pythagoras theorem)

(25–BC)² = (5)² + BC²

⇒ BC = 12

⇒ AC = 13 (by AC = 25 – BC)

By Pythagoras theorem

(25–BC)² = 5² + BC²

⇒ BC = 12

⇒ AC = 13 and AB = 5

we know that

⇒ 

⇒ 

⇒ 

Question 13.

In ΔABC, m∠C = 90 and m∠A = m∠B,

(1) Is cosA = cosB?

(2) Is tanA = tanB?

(3) Will the other trigonometric ratios of ∠A and ∠B be equal?

Answer:

(1) yes, because the side AB and BC will be equal by property of triangle and therefore all the trigonometric ratios of these two angles will always be equal.

(2) yes, because the side AB and BC will be equal by property of triangle and therefore all the trigonometric ratios of these two angles will always be equal.

(3) yes, because the side AB and BC will be equal by property of triangle and therefore all the trigonometric ratios of these two angles will always be equal.

Question 14.

If 3cotA = 4, examine whether  cos²A – sin²A.

Answer:

Let AB = 4, and BC = 3

⇒ AC = 5

we know that

⇒ 

⇒ 

⇒ 

⇒ 

=

R.H.S =

Question 15.

If pcot θ = q, examine whether 

Answer:

L.H.S

dividing by sinθ

⇒ 

substitute cotθ =

⇒ 

⇒ 

= R.H.S

Question 16.

State whether the following are true or false. Justify your answer:

(1) sin θ =  , for some angle having measure θ.

(2) cos θ = , for some angle having measure θ.

(3) cosecA =  for some measure of angle A.

(4) The value of tanA is always less than 1.

(5) secB =  for some ∠B.

(6) cos θ = 100 for some angle having measure θ.

Answer:

(1) No, it is false, because we know that hypotenuse is smallest side and it is the denominator which is smaller than the numerator.

IT IS FALSE.

(2) Yes, it is true, since the denominator is greater than the numerator which implies that it is possible for the hypotenuse to be the greatest side, and ∴ the triangle could be formed.

(3) Yes, it is true because the hypotenuse is the numerator in case of cosec and it is greater than denominator which means the triangle can be formed.

(4) The statement is false.

The value of tan could be greater than 1.

consider a triangle whose tan of an angle is smaller than 1, then the other angle will definitely have a tan greater than 1. because their ratios are just inversed.

(5) The statement is false, in sec ratio numerator is the hypotenuse which is the smaller than the denominator which is not possible.

(6) The Statement is False. Because here numerator is 100 and the denominator is 1, and denominator is the hypotenuse and the numerator is any of the perpendicular side, and hypotenuse is always the longest side. But here it is not so and hence it is not possible.

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Exercise 9.2

Question 1.

Verify:

cos60 = 1 – 2sin²30 = 2cos²30 – 1 = cos²30 – sin²30

Answer:

cos60 =, 

=

2cos²30 – 1 =

=

cos²30 – sin²30

=

=

Question 2.

Verify:

sin60 = 2sin30 cos30

Answer:

cos60 =, 

2sin30 cos30 =

=

Question 3.

Verify:

sin60 = 

Answer:

 and 

L.H.S.

R.H.S

⇒ 

⇒ 

Question 4.

Verify:

cos60 = 

Answer:

cos60 =, 

cos60 =

R.H.S

=

Question 5.

Verify:

cos90 = 4cos³30 – 3cos30

Answer:

R.H.S

= 0

Question 6.

Evaluate:

Answer:

cos60 =, 

=

=

Rationalise the denominator

=

Question 7.

Evaluate:

Answer:

cos60 =, 

denominator =

=

Question 8.

Evaluate:

2sin²30 cot30 – 3cos²60 sec²30

Answer:

=

⇒

Question 9.

Evaluate:

3cos²30 + sec²30 + 2cos0 + 3sin90 – tan²60

Answer:

put all the respective values

=

Question 10.

In ΔABC, m∠B = 90, find the measure of the parts of the triangle other than the ones which are given below:

(1) m∠C = 45, AB = 5

(2) m∠A = 30, AC = 10

(3) AC = 6√2, BC = 3√6

(4) AB = 4, BC = 4

Answer:

(1) If ∠C = 45 and ∠B = 90

⇒ ∠A = 45 (angle sum property of a triangle)

∠A = ∠C

⇒ AB = BC (sides opposite to equal angles)

and since, AB = 5

⇒ BC = 5

and by Pythagoras theorem

AB² + BC² = AC²

⇒ 5² + 5² = AC

⇒ AC = 5√2

(2) if ∠A = 30 and ∠B = 90

⇒ ∠C = 60 (Angle sum property of triangle)

put the known values

also, 

(3)

⇒ ∠C = 30

⇒ ∠A = 60 (by angle sum property)

also, by Pythagoras theorem

AB² = AC² – BC²

⇒ AB = 3√2

(4) Since ∠B = 90

⇒ AB² + BC² = AC²

substituting AC = 4 and BC = 4

we get

AC = 4√2

Also since, AB = BC

⇒ ∠A = ∠C (angles opposite to equal sides of the same triangle)

∵ ∠B = 0

⇒ ∠A = ∠C = 45 (Angle sum property)

Question 11.

In a rectangle ABCD, AB = 20, m∠BAC = 60, calculate the length of side and diagonals and.

Answer:

∵ ∠BAC = 60

we know that

since, diagonals of any rectangle are always equal

⇒ AC = BD

Now

by Pythagoras theorem

AB² + BC² = AC²

⇒ 20² + BC² = 40²

⇒ BC = 20√3

Question 12.

If θ is measure of an acute angle and cosθ = sinθ, find the value of 2tan²θ + sin²θ + 1.

Answer:

cos θ = sin θ

⇒ tan θ = 1

now ∵ θ is an acute angle and tan θ is 1.

⇒ θ = 45

now by substituting 

we get

Question 13.

If α is measure of acute angle and 3sinα = 2cosα, prove that 

Answer:

3sinα = 2cosα

L.H.S

by substituting

= 1

Question 14.

If A = 30 and B = 60, verify that

sin(A + B) = sinA cosB + cosA sinB,

Answer:

L.H.S

= Sin(A + B)

= sin(30 + 60)

= sin90

= 1

R.H.S

substituting the required values in

sinAcosB + cosAsinB

=

= 1

Question 15.

If A = 30 and B = 60, verify that

cos(A + B) = cosA cosB – sinA sinB

Answer:

L.H.S = cos(A + B)

⇒ cos(60 + 30)

⇒ cos90 = 0

R.H.S

cosA cosB – sinA sinB

⇒ cos60 cos30 – sin60 sin30

Question 16.

If sin(A – B) = sinA cosB – cosA sinB and cos(A – B) = cosA cosB + sinA sinB, find the values of sin15 and cos15.

Answer:

substituting A = 45 and B = 30 in

sin(A – B) = sinA cosB – cosA sinB

⇒ sin(45 – 30) = sin45 cos30 – cos45 sin30

Question 17.

State whether the following are true or false. Justify your answer:

(1) The value of sinθ increases as θ increases from 0 to 90.

(2) sinθ = cosθ for all value ofθ.

(3) cos(A + B) = cosA + cosB

(4) tanA is not defined for A = 90.

(5) The value of cot increases as θ increases from 0 to 90.

Answer:

(1) True

sin0 = 0

sin30 =

sin60 =

sin90 = 1

we can see its increasing with increase in the angle in the range 0–90.

(2) False

They are only equal at θ = 45, otherwise are not equal.

(3) False

Let A = B = 45

L.H.S

⇒ cos90 = 0

R.H.S

⇒ cos45 + cos45

(4) True

which is not defined

(5) False

it decreases with increase in θ.

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Exercise 9.3

Question 1.

Evaluate:

Answer:

sin(90–θ ) = cosθ

= 1

Question 2.

Evaluate:

tan48 — cot42

Answer:

cot(90–θ ) = tanθ

tan48 – cot42

= tan48 – cot(90–48)

= tan48 – tan48

= 0

Question 3.

Evaluate:

cosec32 — sec58

Answer:

sec(90–θ ) = cosecθ

cosec32 — sec58

= cosec32 — sec(90–32)

= cosec32 — cosec32

= 0

Question 4.

Evaluate:

 + cos59 ⋅ cosec31

Answer:

sin(90–θ ) = cosθ

⇒ 

⇒ 2

Question 5.

Evaluate:

sec70 sin20 — cos20 cosec70

Answer:

sec(90–θ ) = cosecθ

cosec(90–θ ) = secθ

sec70 sin20 — cos20 cosec70

= sec(90–20)sin20 – cos20cosec(90–20)

= cosec20sin20 – cos20sec20

= 0

Question 6.

Evaluate:

cos(40—) — sin(50 + ) + 

Answer:

sin(90–θ ) = cosθ

cos(90–θ ) = sinθ

= 1

Question 7.

Evaluate:

 + 

Answer:

sin(90–θ ) = cosθ

cos(90–θ ) = sinθ

tan(90–θ ) = cotθ

= 1 + 1

= 2

Question 8.

Evaluate:

cot12 ⋅ cot38 ⋅ cot52 ⋅ cot60 ⋅ cot78

Answer:

cot(90–78).cot(90–52).cot52..cot78

Question 9.

Evaluate:

 + √3 (tan10 tan30 tcm40 tan50 tan80–

Answer:

cos(90–θ ) = sinθ

tan(90–θ ) = cotθ

sin(90–θ ) = cosθ

cot(90–θ ) = tanθ

cosec(90–θ ) = secθ

sec(90–θ) = cosecθ

 2

Question 10.

Evaluate:

Answer:

cos(90–θ ) = sinθ

tan(90–θ ) = cotθ

sin(90–θ ) = cosθ

cot(90–θ ) = tanθ

cosec(90–θ ) = secθ

sec(90–θ) = cosecθ

Question 11.

Prove the following:

sin48 sec42 + cos48 cosec42 = 2

Answer:

cos(90–θ ) = sinθ

tan(90–θ ) = cotθ

sin(90–θ ) = cosθ

cot(90–θ ) = tanθ

cosec(90–θ ) = secθ

sec(90–θ) = cosecθ

L.H.S

sin48 sec42 + cos48 cosec42

= sin48sec(90–48) + cos48cosec(90–48)

= sin48cosec48 + cos48sec48

= 1 + 1

= 2 = R.H.S

Question 12.

Prove the following:

 — 2cos70 cosec20 = 0

Answer:

cos(90–θ ) = sinθ

tan(90–θ ) = cotθ

sin(90–θ ) = cosθ

cot(90–θ ) = tanθ

cosec(90–θ ) = secθ

sec(90–θ) = cosecθ

L.H.S

= 1 + 1–2

= 0 = R.H.S

Question 13.

Prove the following:

Answer:

cos(90–θ ) = sinθ

tan(90–θ ) = cotθ

sin(90–θ ) = cosθ

cot(90–θ ) = tanθ

cosec(90–θ ) = secθ

sec(90–θ) = cosecθ

L.H.S.

= 0

Question 14.

Prove the following:

Answer:

cos(90–θ ) = sinθ

tan(90–θ ) = cotθ

sin(90–θ ) = cosθ

cot(90–θ ) = tanθ

cosec(90–θ ) = secθ

sec(90–θ) = cosecθ

L.H.S

Question 15.

Express the following in terms of trigonometric ratios of angles having measure between 0 and 45:

(1) sin85 + cosec85

(2) cos89 + cosec87

(3) sec81 + cosec54

Answer:

cos(90–θ ) = sinθ

tan(90–θ ) = cotθ

sin(90–θ ) = cosθ

cot(90–θ ) = tanθ

cosec(90–θ ) = secθ

sec(90–θ) = cosecθ

(1) sin85 + cosec85

sin(90–5) + cosec(90–5)

= cos5 + sec5

(2) cos89 + cosec87

cos(90–1) + cosec(90–3)

= sin1 + sec3

(3) sec81 + cosec54

sec(90–9) + cosec(90–36)

= cosec9 + sec36

Question 16.

For ΔABC, prove that (1)  ,

(2) 

Answer:

(a) cos(90–θ ) = sinθ

tan(90–θ ) = cotθ

sin(90–θ ) = cosθ

cot(90–θ ) = tanθ

cosec(90–θ ) = secθ

sec(90–θ) = cosecθ

A + B + C = 180 (Angle sum property of triangle)

⇒ A + C = 180–B

dividing both sides by 2

(2) A + B + C = 180

⇒ B + C = 180–A

dividing by 2

Question 17.

If A + B = 90, prove that 

Answer:

cos(90–θ ) = sinθ

tan(90–θ ) = cotθ

sin(90–θ ) = cosθ

cot(90–θ ) = tanθ

cosec(90–θ ) = secθ

sec(90–θ) = cosecθ

L.H.S

∵ A + B = 90

∵ 

= secA

Question 18.

If 3 θ is the measure of an acute angle and sin30 = cos(θ — 26), then find the value of θ.

Answer:

cos(90–θ ) = sinθ

tan(90–θ ) = cotθ

sin(90–θ ) = cosθ

cot(90–θ ) = tanθ

cosec(90–θ ) = secθ

sec(90–θ) = cosecθ

sin30 = cos(90–30)

⇒ cos(90–30) = cos(θ –26)

⇒ 90–30 = θ–26

⇒ θ = 86

Question 19.

If 0 < θ < 90, θ, sinθ = cos30, then obtain the value of 2tan²θ — 1.

Answer:

cos(90–θ ) = sinθ

tan(90–θ ) = cotθ

sin(90–θ ) = cosθ

cot(90–θ ) = tanθ

cosec(90–θ ) = secθ

sec(90–θ) = cosecθ

sinθ = cos30

⇒ θ = 60

Now, 2tan²θ – 1

= 2tan²60 – 1

= 2(√3)² – 1

= 5

Question 20.

If tanA = cotB, prove that A + B = 90, where A and B are measures of acute angles.

Answer:

tan(90–θ ) = cotθ

tanA = cotB

⇒ tanA = tan(90–B)

⇒ A = 90–B

⇒ A + B = 90

Question 21.

If sec2A = cosec(A — 42), where 2A is the measure of an acute angle, find the value of A.

Answer:

sec(90–θ) = cosecθ

sec2A = cosec(A–42)

∴ sec2A = sec(90–(A–42))

⇒ 2A = 90 – A + 42

⇒ A = 44

Question 22.

If 0 < θ < 90 and secθ = cosec60, find the value of 2cos²θ — 1.

Answer:

secθ = cosec60

⇒ secθ =

⇒ θ = 30

Now,

2cos²— 1

= 2cos²30— 1

---

Exercise 9

Question 1.

Prove the following by using trigonometric identities:

Answer:

1 + cot²θ = cosec²θ

= 1

Question 2.

Prove the following by using trigonometric identities:

2sin²θ + 4sec²θ + 5cot²θ + 2cos²θ — 4tan²θ — 5cosec²θ = 1

Answer:

L.H.S

2sin²θ + 4sec²θ + 5cot²θ + 2cos²θ — 4tan²θ — 5cosec²θ

= (2sin²θ + 2cos²θ) + ( 4sec²θ – 4tan²) + (5cot²θ — 5cosec²θ)

= 2 + 4 – 5

= 1

Question 3.

Prove the following by using trigonometric identities:

Answer:

L.H.S

Question 4.

Prove the following by using trigonometric identities:

Answer:

L.H.S

Question 5.

Prove the following by using trigonometric identities:

Answer:

L.H.S

Question 6.

Prove the following by using trigonometric identities:

Answer:

L.H.S

Question 7.

Prove the following by using trigonometric identities:

Answer:

L.H.S

Question 8.

Prove the following by using trigonometric identities:

(sinθ + cosecθ)² + (cosθ + secθ)² = 7 + tan²θ + cot²θ.

Answer:

L.H.S

(sinθ + cosecθ)² + (cosθ + secθ)²

= sin²θ + 2sinθcosecθ + cosec²θ + cos²θ + 2cosθsecθ + sec²θ

= sin²θ + 2 + cosec²θ + cos²θ + 2 + sec²θ

= sin²θ + cos²θ + sec²θ + cosec²θ + 2 + 2

= 1 + 2 + 2 + (1 + tan²θ) + (1 + cot²θ)

= 7 + tan²θ + cot²θ

Question 9.

Prove the following by using trigonometric identities:

2sec²θ — sec⁴θ — 2cosec²θ + cosec⁴θ = cot⁴θ — tan⁴θ.

Answer:

2sec²θ — sec⁴θ — 2cosec²θ + cosec⁴θ

= 2(1 + tan²θ) – (1 + tan²θ)² – 2(1 + cot²θ) + (1 + cot²θ)²

open all the brackets and cancel terms

= cot⁴θ – tan⁴θ

Question 10.

Prove the following by using trigonometric identities:

(sinθ — secθ)² + (cosθ — cosecθ)² = (1 — secθ ⋅ cosecθ)².

Answer:

L.H.S

= sin²θ – 2sinθsecθ + sec²θ + cos²θ – 2cosθcosecθ + cosec²θ

= (sin²θ + cos²θ) – (2sinθsecθ + 2cosθcosecθ) + cosec²θ + sec²θ

Question 11.

Prove the following by using trigonometric identities:

Answer:

L.H.S

Question 12.

Prove the following by using trigonometric identities:

Answer:

Question 13.

Prove the following by using trigonometric identities:

Answer:

L.H.S

Question 14.

Prove the following by using trigonometric identities:

Answer:

L.H.S

∵ tanθcotθ = 1

Question 15.

Prove the following by using trigonometric identities:

Answer:

∵ 

Question 16.

Prove the following by using trigonometric identities:

Answer:

Question 17.

Prove the following by using trigonometric identities:

sin⁴θ – cos⁴θ = sin²θ – cos²θ = 2sin²θ – 1 = 1 – 2 cos²θ.

Answer:

 . . .(1)

Now, 

Also, 

By (1)

Question 18.

Prove the following by using trigonometric identities:

tan²A — tan²B = 

Answer:

Question 19.

Prove the following by using trigonometric identities:

2(sin⁶θ + cos⁶θ) — 3(sin⁴θ + cos⁴θ) + 1 = 0

Answer:

2((sin²θ)³ + (cos²θ)³) — 3(sin⁴θ + cos⁴θ) + 1

(∵ a² + b² = (a + b)²–2ab)

⇒ 2[(sin²θ + cos²θ)(sin⁴θ –sin²θcos²θ + cos⁴θ)]–3[(sin²θ + cos²θ)–2sin²θcos²θ] + 1

= 2[(sin⁴θ –sin²θcos²θ + cos⁴θ)]–3[1–2sin²θcos²θ] + 1

= 2sin⁴θ–2sin²θcos²θ + 2cos⁴θ–3 + 6sin²θcos²θ + 1

= 2(sin⁴θ + 4sin²θcos²θ + cos⁴θ)–2

= 2(sin²θ + cos²θ)²–2

= 2–2

= 0

Question 20.

If sinθ + cosθ = p and secθ + cosecθ = q, show that q(p² — 1) = 2p.

Answer:

L.H.S

q(p² — 1)

substituting p and q.

(sec + cosec)((sin + cos)²–1)

=

also 

Question 21.

If tanθ + sin = a and tanθ — sinθ = b, then prove that a² — b² = 

Answer:

tan + sin = a [1]

tan — sin = b [2]

adding and subtracting [1] and [2]

2tanθ = a + b and 2sinθ = a–b

L.H.S

a² — b² = (a + b)(a–b)

⇒ a² — b² = (2tanθ)( 2sinθ)

⇒ a² — b² = 4tanθsinθ

R.H.S

4√(ab)

⇒ 

=

L.H.S. = R.H.S

Question 22.

acosθ + bsinθ = p and asinθ— bcosθ = q, then prove that a² + b² = p² + q².

Answer:

acosθ + bsinθ = p (i)

asinθ – bcosθ = q (ii)

Squaring and adding (i) and (ii)

∴ (acosθ + bsinθ)² + (asinθ – bcosθ)² = p² + q²

∴ a²cos²θ + 2abcosθsinθ + b²sin²θ + a²sin²θ – 2absinθcosθ + b²cos²θ = p² + q²

∴ a²(cos²θ + sin²θ) + b² (sin²θ + cos²θ) = p² + q²

∴a² (1) + b² (1) = p² + q²

∴ a² + b² = p² + q²

Question 23.

secθ + tanθ = p, then obtain the values of secθ, tanθ and sinθ in terms of p.

Answer:

secθ + tanθ [1]

sec²θ –tan²θ = 1

⇒ (secθ – tanθ)( (secθ + tanθ)) = 1

⇒ (secθ – tanθ) p = 1

⇒ (secθ – tanθ) = [2]

Adding and subtracting [1] and [2]

 and 

⇒  and 

⇒  and 

Question 24.

Evaluate the following:

 ⋅ tan17 tan38 tan60 tan52 tan73 — 3(sin²32 + sin²58)

Answer:

Question 25.

Evaluate the following:

Answer:

Question 26.

If sinA + cosA = √2 sin(90—A), then obtain the value of cotA.

Answer:

sinA + cosA = sin(90—A)

dividing the complete equation by cosA

Question 27.

If cosecθ = √2, then find the value of 

Answer:

cosecθ = √2

Now, 

Question 28.

If  then evaluate 

Answer:

1 + tan²θ = sec²θ

⇒ 

and 

Now, 

Question 29.

If , 0 < θ < 90, find the value of sinθ and tanθ.

Answer:

sin²θ = 1 – cos²θ

∴ 

Question 30.

If θ is the measure of an acute angle such that bsinθ = acosθ, then  is =
A. 

B. 

C. 

D. 

Answer:

dividing both numerator and denominator by cosθ.

Question 31.

Which of the following is correct for some 0 such that 0 ≤ θ < 90?
A.  >1

B.  = 1

C. sec θ = 0

D. <1

Answer:

for θ = 0, secθ = 1

⇒ option B is correct

Question 32.

If , then  is ….
A. 

B. 

C. 

D. 3

Answer:

Question 33.

If , then the value of  is ____
A. 

B. 

C. 

D. 

Answer:

⇒ 

∵ tan²θ =

Question 34.

If , then the value of  is ____
A. 7

B. 

C. 

D. 

Answer:

dividing numerator and denominator by sinθ

Question 35.

If cosecA =  and A + B = 90, then secB is ………..
A. 

B. 

C. 

D. 

Answer:

∵ A = 90 – B

Question 36.

If θ is the measure of an acute angle and √3 sinθ = cosθ, then θ is ____
A. 30

B. 45

C. 60

D. 90

Answer:

Question 37.

If  then the value of (sinA + cosA) secA is …..
A. 

B. 

C. 

D. 

Answer:

(sinA + cosA) secA

Question 38.

If then the value of is ….
A. 

B. 3

C. 

D. 

Answer:

0

sinA = cosA.tanA

Now, 

Question 39.

In ΔABC, if m∠ABC = 90, m∠ACB = 45 and AC = 6, then area of ΔABC is …..
A. 18

B. 36

C. 9

D. 

Answer:

given ∠B = 90, ∠C = 45

by angle sum property

⇒ ∠A = 45

⇒ AB = BC

also AB² + BC² = AC² (by (Pythagoras theorem)

⇒ AB√2 = AC

∵ AC = 6

⇒ AB =

⇒ AB = BC = 3√2

Area of triangle =

⇒ Area =

= 9

Question 40.

If cos²45 – cos²30 = x ⋅ cos45 ⋅ sin45, then x is …….
A. 2

B. 

C. 

D. 

Answer:

cos²45 – cos²30 = x ⋅ cos45 ⋅ sin45

Question 41.

If A and B are complementary angles, then sinA ⋅ secB is ____
A. 1

B. 0

C. —1

D. 2

Answer:

Question 42.

The value of tan20 tan25 tan45 tan65 tan70 is ………….
A. —1

B. 1

C. 0

D. √3

Answer:

tan20.tan25.tan45.tan65.tan70

= cot(90–20).cot(90–25).1.tan65tan70

= cot70.cot65.tan65.tan70

= 1

Question 43.

If 7θ and 2θ are measure of acute angles such that sin7θ = cos2θ then 2sin3θ — √3 tan3θ is ……….
A. 1

B. 0

C. —1

D. 1 — √3

Answer:

sin7θ = cos2θ

∴ sin7θ = sin(90–2θ )

⇒ 9θ = 90

⇒ θ = 10

2sin3θ –√3tan3θ

put θ = 10

2sin3×20 – √3×tan (10× 3)

= 0

Question 44.

If A + B = 90, then  is ….
A. cot²B

B. tan²A

C. cot²A

D. —cot²A

Answer:

Question 45.

For ΔABC, sin = ……..
A. sin

B. sinA

C. cos

D. cosA

Answer:

Question 46.

 = ………..
A. 1

B. 2

C. 3

D. 0

Answer:

Question 47.

If 7cos²θ + 3sin²θ = 4, then cotθ is ___
A. 7

B. 

C. 

D. 

Answer:

7cos² + 3sin² = 4

⇒ 7(1–sin²θ) + 3sin² = 4

⇒ 7–4sin²θ = 4

⇒ sinθ =

⇒ θ = 60

⇒ cot60 =

Question 48.

If tan5θ ⋅ tan4θ = 1, θ is ____
A. 7

B. 3

C. 10

D. 9

Answer:

tan5θ.cot(90–4θ) = 1

⇒ cot(90–4θ) = tan5θ.

⇒ 5θ = 90– 4θ

⇒ θ = 10

Question 49.

If A and B are measures of acute angles and tanA = and sinB = , then cos(A + B) is ………………..
A. 0

B. 

C. 

D. 

Answer:

tanA =

also tan30 =

⇒ A = 30

Sin B =

⇒ B = 30

⇒ A + B = 30 + 30 = 60

⇒ cos(A + B) = Cos(60)