In ΔABC, m∠A = 90. If AB = 5, AC = 12 and BC = 13, find sinC, cosC, tanB, cosB, sinB.
we know that
In ΔABC, m∠B = 90. If BC = 3 and AC = 5, find all the six trigonometric ratios of ∠A.
Pythagoras theorem
32 + AB2 = 52
⇒ AB = 4
we know that
If cosA = find sinA and tanA.
Let AB = 4 and AC = 5
BC = 3 (by Pythagoras theorem)
we know that
⇒
⇒
If cosec θ = find tan θ and cos θ.
Let ∠A be θ
BC = 5k, AC = 13k
⇒ AB = 12k (by Pythagoras theorem)
we know that
⇒
⇒
If cosB = , find the other five trigonometric ratios.
Let BC = k and AB = 3k
⇒ AC = (Pythagoras theorem)
we know that
In ΔABC, m∠A = 90 and if AB : BC = 1 : 2 find sinB, cosC, tanC.
AB = k and BC = 2k
AC = (by Pythagoras theorem)
If tan θ = , find the value of
Let’s divide both numerator and denominator by cosθ, we get
put tanθ = in this equation.
If sec θ = find the value of
Let ∠B = θ
⇒ AB = 5k, BC = 13k
⇒ AC = 12k (by Pythagoras theorem)
we know that
⇒
⇒
putting the above values in the given equations.
=
If sinB = prove that 3cosB – 4cos3B = 0.
Let AC = 1 and AB = 2.
⇒
∴ BC = √3 (by Pythagoras theorem) we know that
⇒
⇒
=
= 0
which is equal to the R.H.S
If tanA = √3, verify that
(1) sin2A + cos2A = 1
(2) sec2A – tan2A = 1
(3) 1 + cot2A = cosec2A
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
If cos θ = , verify that tan2θ – sin2θ = tan2θ⋅ sin2θ
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
In ΔABC, m∠B = 90, AC + BC = 25 and AB = 5, determine the value of sinA, cosA and tanA.
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
⇒
⇒
⇒
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?
(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.
If 3cotA = 4, examine whether cos2A – sin2A.
Let AB = 4, and BC = 3
⇒ AC = 5
we know that
⇒
⇒
⇒
⇒
=
R.H.S =
If pcot θ = q, examine whether
L.H.S
dividing by sinθ
⇒
substitute cotθ =
⇒
⇒
= R.H.S
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.
(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.
Verify:
cos60 = 1 – 2sin230 = 2cos230 – 1 = cos230 – sin230
cos60 =,
=
2cos230 – 1 =
=
cos230 – sin230
=
=
Verify:
sin60 = 2sin30 cos30
cos60 =,
2sin30 cos30 =
=
Verify:
sin60 =
and
L.H.S.
R.H.S
⇒
⇒
Verify:
cos60 =
cos60 =,
cos60 =
R.H.S
=
Verify:
cos90 = 4cos330 – 3cos30
R.H.S
= 0
Evaluate:
cos60 =,
=
=
Rationalise the denominator
=
Evaluate:
cos60 =,
denominator =
=
Evaluate:
2sin230 cot30 – 3cos260 sec230
=
⇒
Evaluate:
3cos230 + sec230 + 2cos0 + 3sin90 – tan260
put all the respective values
=
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
(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)
In a rectangle ABCD, AB = 20, m∠BAC = 60, calculate the length of side and diagonals and.
∵ ∠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
If θ is measure of an acute angle and cosθ = sinθ, find the value of 2tan2θ + sin2θ + 1.
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
If α is measure of acute angle and 3sinα = 2cosα, prove that
3sinα = 2cosα
L.H.S
by substituting
= 1
If A = 30 and B = 60, verify that
sin(A + B) = sinA cosB + cosA sinB,
L.H.S
= Sin(A + B)
= sin(30 + 60)
= sin90
= 1
R.H.S
substituting the required values in
sinAcosB + cosAsinB
=
= 1
If A = 30 and B = 60, verify that
cos(A + B) = cosA cosB – sinA sinB
L.H.S = cos(A + B)
⇒ cos(60 + 30)
⇒ cos90 = 0
R.H.S
cosA cosB – sinA sinB
⇒ cos60 cos30 – sin60 sin30
If sin(A – B) = sinA cosB – cosA sinB and cos(A – B) = cosA cosB + sinA sinB, find the values of sin15 and cos15.
substituting A = 45 and B = 30 in
sin(A – B) = sinA cosB – cosA sinB
⇒ sin(45 – 30) = sin45 cos30 – cos45 sin30
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.
(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.
Evaluate:
sin(90–θ ) = cosθ
= 1
Evaluate:
tan48 — cot42
cot(90–θ ) = tanθ
tan48 – cot42
= tan48 – cot(90–48)
= tan48 – tan48
= 0
Evaluate:
cosec32 — sec58
sec(90–θ ) = cosecθ
cosec32 — sec58
= cosec32 — sec(90–32)
= cosec32 — cosec32
= 0
Evaluate:
+ cos59 ⋅ cosec31
sin(90–θ ) = cosθ
⇒
⇒ 2
Evaluate:
sec70 sin20 — cos20 cosec70
sec(90–θ ) = cosecθ
cosec(90–θ ) = secθ
sec70 sin20 — cos20 cosec70
= sec(90–20)sin20 – cos20cosec(90–20)
= cosec20sin20 – cos20sec20
= 0
Evaluate:
cos(40—) — sin(50 + ) +
sin(90–θ ) = cosθ
cos(90–θ ) = sinθ
= 1
Evaluate:
+
sin(90–θ ) = cosθ
cos(90–θ ) = sinθ
tan(90–θ ) = cotθ
= 1 + 1
= 2
Evaluate:
cot12 ⋅ cot38 ⋅ cot52 ⋅ cot60 ⋅ cot78
cot(90–78).cot(90–52).cot52..cot78
Evaluate:
+ √3 (tan10 tan30 tcm40 tan50 tan80–
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
Evaluate:
cos(90–θ ) = sinθ
tan(90–θ ) = cotθ
sin(90–θ ) = cosθ
cot(90–θ ) = tanθ
cosec(90–θ ) = secθ
sec(90–θ) = cosecθ
Prove the following:
sin48 sec42 + cos48 cosec42 = 2
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
Prove the following:
— 2cos70 cosec20 = 0
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
Prove the following:
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
Prove the following:
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
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
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
(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
If A + B = 90, prove that
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
If 3 θ is the measure of an acute angle and sin30 = cos(θ — 26), then find the value of θ.
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
If 0 < θ < 90, θ, sinθ = cos30, then obtain the value of 2tan2θ — 1.
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
If tanA = cotB, prove that A + B = 90, where A and B are measures of acute angles.
tan(90–θ ) = cotθ
tanA = cotB
⇒ tanA = tan(90–B)
⇒ A = 90–B
⇒ A + B = 90
If sec2A = cosec(A — 42), where 2A is the measure of an acute angle, find the value of A.
sec(90–θ) = cosecθ
sec2A = cosec(A–42)
∴ sec2A = sec(90–(A–42))
⇒ 2A = 90 – A + 42
⇒ A = 44
If 0 < θ < 90 and secθ = cosec60, find the value of 2cos2θ — 1.
secθ = cosec60
⇒ secθ =
⇒ θ = 30
Now,
2cos2— 1
= 2cos230— 1
Prove the following by using trigonometric identities:
1 + cot2θ = cosec2θ
= 1
Prove the following by using trigonometric identities:
2sin2θ + 4sec2θ + 5cot2θ + 2cos2θ — 4tan2θ — 5cosec2θ = 1
L.H.S
2sin2θ + 4sec2θ + 5cot2θ + 2cos2θ — 4tan2θ — 5cosec2θ
= (2sin2θ + 2cos2θ) + ( 4sec2θ – 4tan2) + (5cot2θ — 5cosec2θ)
= 2 + 4 – 5
= 1
Prove the following by using trigonometric identities:
L.H.S
Prove the following by using trigonometric identities:
L.H.S
Prove the following by using trigonometric identities:
L.H.S
Prove the following by using trigonometric identities:
L.H.S
Prove the following by using trigonometric identities:
L.H.S
Prove the following by using trigonometric identities:
(sinθ + cosecθ)2 + (cosθ + secθ)2 = 7 + tan2θ + cot2θ.
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
Prove the following by using trigonometric identities:
2sec2θ — sec4θ — 2cosec2θ + cosec4θ = cot4θ — tan4θ.
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
Prove the following by using trigonometric identities:
(sinθ — secθ)2 + (cosθ — cosecθ)2 = (1 — secθ ⋅ cosecθ)2.
L.H.S
= sin2θ – 2sinθsecθ + sec2θ + cos2θ – 2cosθcosecθ + cosec2θ
= (sin2θ + cos2θ) – (2sinθsecθ + 2cosθcosecθ) + cosec2θ + sec2θ
Prove the following by using trigonometric identities:
L.H.S
Prove the following by using trigonometric identities:
Prove the following by using trigonometric identities:
L.H.S
Prove the following by using trigonometric identities:
L.H.S
∵ tanθcotθ = 1
Prove the following by using trigonometric identities:
∵
Prove the following by using trigonometric identities:
Prove the following by using trigonometric identities:
sin4θ – cos4θ = sin2θ – cos2θ = 2sin2θ – 1 = 1 – 2 cos2θ.
. . .(1)
Now,
Also,
By (1)
Prove the following by using trigonometric identities:
tan2A — tan2B =
Prove the following by using trigonometric identities:
2(sin6θ + cos6θ) — 3(sin4θ + cos4θ) + 1 = 0
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
If sinθ + cosθ = p and secθ + cosecθ = q, show that q(p2 — 1) = 2p.
L.H.S
q(p2 — 1)
substituting p and q.
(sec + cosec)((sin + cos)2–1)
=
also
If tanθ + sin = a and tanθ — sinθ = b, then prove that a2 — b2 =
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
acosθ + bsinθ = p and asinθ— bcosθ = q, then prove that a2 + b2 = p2 + q2.
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
secθ + tanθ = p, then obtain the values of secθ, tanθ and sinθ in terms of p.
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
Evaluate the following:
⋅ tan17 tan38 tan60 tan52 tan73 — 3(sin232 + sin258)
Evaluate the following:
If sinA + cosA = √2 sin(90—A), then obtain the value of cotA.
sinA + cosA = sin(90—A)
dividing the complete equation by cosA
If cosecθ = √2, then find the value of
cosecθ = √2
Now,
If then evaluate
1 + tan2θ = sec2θ
⇒
and
Now,
If , 0 < θ < 90, find the value of sinθ and tanθ.
sin2θ = 1 – cos2θ
∴
If θ is the measure of an acute angle such that bsinθ = acosθ, then is =
A.
B.
C.
D.
dividing both numerator and denominator by cosθ.
Which of the following is correct for some 0 such that 0 ≤ θ < 90?
A. >1
B. = 1
C. sec θ = 0
D. <1
for θ = 0, secθ = 1
⇒ option B is correct
If , then is ….
A.
B.
C.
D. 3
If , then the value of is ____
A.
B.
C.
D.
⇒
∵ tan2θ =
If , then the value of is ____
A. 7
B.
C.
D.
dividing numerator and denominator by sinθ
If cosecA = and A + B = 90, then secB is ………..
A.
B.
C.
D.
∵ A = 90 – B
If θ is the measure of an acute angle and √3 sinθ = cosθ, then θ is ____
A. 30
B. 45
C. 60
D. 90
If then the value of (sinA + cosA) secA is …..
A.
B.
C.
D.
(sinA + cosA) secA
If then the value of is ….
A.
B. 3
C.
D.
0
sinA = cosA.tanA
Now,
In ΔABC, if m∠ABC = 90, m∠ACB = 45 and AC = 6, then area of ΔABC is …..
A. 18
B. 36
C. 9
D.
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
If cos245 – cos230 = x ⋅ cos45 ⋅ sin45, then x is …….
A. 2
B.
C.
D.
cos245 – cos230 = x ⋅ cos45 ⋅ sin45
If A and B are complementary angles, then sinA ⋅ secB is ____
A. 1
B. 0
C. —1
D. 2
The value of tan20 tan25 tan45 tan65 tan70 is ………….
A. —1
B. 1
C. 0
D. √3
tan20.tan25.tan45.tan65.tan70
= cot(90–20).cot(90–25).1.tan65tan70
= cot70.cot65.tan65.tan70
= 1
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
sin7θ = cos2θ
∴ sin7θ = sin(90–2θ )
⇒ 9θ = 90
⇒ θ = 10
2sin3θ –√3tan3θ
put θ = 10
2sin3×20 – √3×tan (10× 3)
= 0
If A + B = 90, then is ….
A. cot2B
B. tan2A
C. cot2A
D. —cot2A
For ΔABC, sin = ……..
A. sin
B. sinA
C. cos
D. cosA
= ………..
A. 1
B. 2
C. 3
D. 0
If 7cos2θ + 3sin2θ = 4, then cotθ is ___
A. 7
B.
C.
D.
7cos2 + 3sin2 = 4
⇒ 7(1–sin2θ) + 3sin2 = 4
⇒ 7–4sin2θ = 4
⇒ sinθ =
⇒ θ = 60
⇒ cot60 =
If tan5θ ⋅ tan4θ = 1, θ is ____
A. 7
B. 3
C. 10
D. 9
tan5θ.cot(90–4θ) = 1
⇒ cot(90–4θ) = tan5θ.
⇒ 5θ = 90– 4θ
⇒ θ = 10
If A and B are measures of acute angles and tanA = and sinB = , then cos(A + B) is ………………..
A. 0
B.
C.
D.
tanA =
also tan30 =
⇒ A = 30
Sin B =
⇒ B = 30
⇒ A + B = 30 + 30 = 60
⇒ cos(A + B) = Cos(60)