Trigonometry
Class 10th Mathematics Gujarat Board Solution
- In ΔABC, m∠A = 90. If AB = 5, AC = 12 and BC = 13, find sinC, cosC, tanB, cosB,…
- In ΔABC, m∠B = 90. If BC = 3 and AC = 5, find all the six trigonometric ratios…
- If cosA = 4/5 find sinA and tanA.
- If cosec θ = 13/5 find tan θ and cos θ.
- If cosB = 1/3 , find the other five trigonometric ratios.
- In ΔABC, m∠A = 90 and if AB : BC = 1 : 2 find sinB, cosC, tanC.
- If tan θ = 4/3 , find the value of 5sintegrate heta +2costheta /3sintegrate heta…
- If sec θ = 13/5 find the value of 5sintegrate heta +3costheta /5costheta…
- If sinB = 1/2 prove that 3cosB - 4cos^3 B = 0.
- If tanA = √3, verify that (1) sin^2 A + cos^2 A = 1 (2) sec^2 A - tan^2 A = 1…
- If cos θ = 2 root 2/3 , verify that tan^2 θ - sin^2 θ = tan^2 θ⋅ sin^2 θ…
- In ΔABC, m∠B = 90, AC + BC = 25 and AB = 5, determine the value of sinA, cosA…
- In ΔABC, m∠C = 90 and m∠A = m∠B, (1) Is cosA = cosB? (2) Is tanA = tanB? (3)…
- If 3cotA = 4, examine whether 1-tan^2a/1+tan^2a cos^2 A - sin^2 A.…
- If pcot θ = q, examine whether psi ntheta -qcostheta /psi ntheta +qcostheta =…
- State whether the following are true or false. Justify your answer: (1) sin θ =…
- cos60 = 1 - 2sin^2 30 = 2cos^2 30 - 1 = cos^2 30 - sin^2 30 Verify:…
- sin60 = 2sin30 cos30 Verify:
- sin60 = 2tan30/1+tan^230 Verify:
- cos60 = 1-tan^230/1+tan^230 Verify:
- cos90 = 4cos^3 30 - 3cos30 Verify:
- sin30+tan45-cosec60/sec30+cos60+cot45 Evaluate:
- 5cos^260+4sec^230-tan^245/sin^230+cos^230 Evaluate:
- 2sin^2 30 cot30 - 3cos^2 60 sec^2 30 Evaluate:
- 3cos^2 30 + sec^2 30 + 2cos0 + 3sin90 - tan^2 60 Evaluate:
- In ΔABC, m∠B = 90, find the measure of the parts of the triangle other than the…
- In a rectangle ABCD, AB = 20, m∠BAC = 60, calculate the length of side bar bc…
- If θ is measure of an acute angle and cosθ = sinθ, find the value of 2tan^2 θ +…
- If α is measure of acute angle and 3sinα = 2cosα, prove that (1-tan^2alpha…
- sin(A + B) = sinA cosB + cosA sinB, If A = 30 and B = 60, verify that…
- cos(A + B) = cosA cosB - sinA sinB If A = 30 and B = 60, verify that…
- If sin(A - B) = sinA cosB - cosA sinB and cos(A - B) = cosA cosB + sinA sinB,…
- State whether the following are true or false. Justify your answer: (1) The…
- cos18/sin72 Evaluate:
- tan48 — cot42 Evaluate:
- cosec32 — sec58 Evaluate:
- cos70/sin20 + cos59 ⋅ cosec31 Evaluate:
- sec70 sin20 — cos20 cosec70 Evaluate:
- cos(40— theta) — sin(50 + theta) + cos^240+cos^250/sin^240+sin^250 Evaluate:…
- cos70/sin20 + cos55cosec35/tan5tan25tan45tan65tan85 Evaluate:
- cot12 ⋅ cot38 ⋅ cot52 ⋅ cot60 ⋅ cot78 Evaluate:
- sin18/cos72 + √3 (tan10 tan30 tcm40 tan50 tan80- Evaluate:
- cos70/sin20 + sin22/cos68 - cos38cosec52/tan18tan35tan60tan72tan55 Evaluate:…
- sin48 sec42 + cos48 cosec42 = 2 Prove the following:
- sin70/cos20 + cos6c^20/sec70 — 2cos70 cosec20 = 0 Prove the following:…
- Prove the following:
- cos (90-a) sin (90-a)/tan (90 - phi) = sin^2a Prove the following:…
- Express the following in terms of trigonometric ratios of angles having measure…
- For ΔABC, prove that (1) tan (a+c/2) = cot b/2 , (2) cos (b+c/2) = sin (a/2)…
- If A + B = 90, prove that root tanatanb+tanacotb/sinasecb = seca
- If 3 θ is the measure of an acute angle and sin30 = cos(θ — 26), then find the…
- If 0 θ 90, θ, sinθ = cos30, then obtain the value of 2tan^2 θ — 1.…
- If tanA = cotB, prove that A + B = 90, where A and B are measures of acute…
- If sec2A = cosec(A — 42), where 2A is the measure of an acute angle, find the…
- If 0 θ 90 and secθ = cosec60, find the value of 2cos^2 θ — 1.
- cos^2theta + 1/1+cot^2theta = 1 Prove the following by using trigonometric…
- 2sin^2 θ + 4sec^2 θ + 5cot^2 θ + 2cos^2 θ — 4tan^2 θ — 5cosec^2 θ = 1 Prove the…
- 1/1+costheta + 1/1-costheta = 2costheta c^2theta Prove the following by using…
- Prove the following by using trigonometric identities:
- root 1-sintegrate heta /1+sintegrate heta = sectheta -tantheta Prove the…
- sectheta +tantheta /costheta c theta +cottheta = costheta c theta -cottheta…
- cottheta +cosectheta -1/cottheta -cosectheta +1 = cosectheta +cottheta Prove the…
- (sinθ + cosecθ)^2 + (cosθ + secθ)^2 = 7 + tan^2 θ + cot^2 θ. Prove the following…
- 2sec^2 θ — sec^4 θ — 2cosec^2 θ + cosec^4 θ = cot^4 θ — tan^4 θ. Prove the…
- (sinθ — secθ)^2 + (cosθ — cosecθ)^2 = (1 — secθ ⋅ cosecθ)^2 . Prove the…
- Prove the following by using trigonometric identities:
- tantheta -cottheta /sintegrate heta costheta = sec^2theta -costheta c^2theta =…
- sectheta -tantheta /sectheta +tantheta = 1-2sectheta tantheta +tan^2theta Prove…
- root sec^2theta +costheta c^3theta = tantheta +cottheta Prove the following by…
- Prove the following by using trigonometric identities:
- tantheta /1-cottheta + cottheta /1-tantheta = 1+tantheta +cottheta = 1+sectheta…
- sin^4 θ - cos^4 θ = sin^2 θ - cos^2 θ = 2sin^2 θ - 1 = 1 - 2 cos^2 θ. Prove the…
- tan^2 A — tan^2 B = Prove the following by using trigonometric identities:…
- 2(sin^6 θ + cos^6 θ) — 3(sin^4 θ + cos^4 θ) + 1 = 0 Prove the following by…
- If sinθ + cosθ = p and secθ + cosecθ = q, show that q(p^2 — 1) = 2p.…
- If tanθ + sin = a and tanθ — sinθ = b, then prove that a^2 — b^2 = 4 root ab…
- acosθ + bsinθ = p and asinθ— bcosθ = q, then prove that a^2 + b^2 = p^2 + q^2 .…
- secθ + tanθ = p, then obtain the values of secθ, tanθ and sinθ in terms of p.…
- sec38/cosec52 + 2/root 3 ⋅ tan17 tan38 tan60 tan52 tan73 — 3(sin^2 32 + sin^2…
- - cottan (90 - theta) + cos6ctheta sectheta (90 - theta)
- If sinA + cosA = √2 sin(90—A), then obtain the value of cotA.
- If cosecθ = √2, then find the value of 2sin^2theta +3cot^2theta /4tan^2theta…
- If tantheta = 8/15 then evaluate (1+sintegrate heta) (2-2sintegrate…
- If costheta = b/root a^2 + b^2 , 0 θ 90, find the value of sinθ and tanθ.…
- If θ is the measure of an acute angle such that bsinθ = acosθ, then…
- Which of the following is correct for some 0 such that 0 ≤ θ 90?A. 1/sectheta…
- If tantheta = 1/root 5 , then cosec^2theta -sec^2theta /cosec^2theta…
- If tan^2theta = 8/7 , then the value of (1+sintegrate heta) (1-sintegrate…
- If cottheta = 4/3 , then the value of costheta -sintegrate heta /costheta…
- If cosecA = 4/3 and A + B = 90, then secB is ………..A. 3/4 B. 1/3 C. 4/3 D. 7/3…
- If θ is the measure of an acute angle and √3 sinθ = cosθ, then θ is ____A. 30…
- If tana = 5/12 then the value of (sinA + cosA) secA is …..A. 12/5 B. 7/12 C.…
- If tantheta = 4/3 then the value of root 1-sintegrate heta /1+sintegrate heta…
- In ΔABC, if m∠ABC = 90, m∠ACB = 45 and AC = 6, then area of ΔABC is …..A. 18…
- If cos^2 45 - cos^2 30 = x ⋅ cos45 ⋅ sin45, then x is …….A. 2 B. 3/2 C. - 1/2…
- If A and B are complementary angles, then sinA ⋅ secB is ____A. 1 B. 0 C. —1…
- The value of tan20 tan25 tan45 tan65 tan70 is ………….A. —1 B. 1 C. 0 D. √3…
- If 7θ and 2θ are measure of acute angles such that sin7θ = cos2θ then 2sin3θ —…
- If A + B = 90, then cotacotb+cotatanb/sina secb - sin^2b/cos^2b is ….A. cot^2…
- For ΔABC, sin (b+c/2) = ……..A. sin a/2 B. sinA C. cos a/2 D. cosA…
- sin^4theta -cos^4theta /sin^2theta -cos^2theta = ………..A. 1 B. 2 C. 3 D. 0…
- If 7cos^2 θ + 3sin^2 θ = 4, then cotθ is ___A. 7 B. 7/3 C. root 3 D. 1/root 3…
- If tan5θ ⋅ tan4θ = 1, θ is ____A. 7 B. 3 C. 10 D. 9
- If A and B are measures of acute angles and tanA = 1/root 3 and sinB = 1/2 ,…
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.
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.
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)