Q1: It is given that tan (θ_{1} + θ_{2}) = where θ_{1} and θ_{2} are acute angles.
Calculate θ_{1} + θ_{2} when tan θ_{1}
Sol:
Now, tan (θ_{1} + θ_{2}) = 1 ⇒ θ_{1} + θ_{2} = 45°.
Q2: Prove that:
Sol:
using tanθ=sinθ/cosθ and cos^{2}θ=1sin^{2}θ as sin^{2}θ+cos^{2}θ=1
Q3: Prove that: (sin^{4} θ – cos^{4} θ +1) cosec^{2} θ = 2
Sol: L.H.S.
= (sin^{4} θ – cos^{4} θ + 1) cosec^{2} θ
= [(sin^{2} θ)^{2} – (cos^{2} θ)^{2} + 1] cosec^{2} θ as [a^{2}b^{2}=(ab)(a+b)]
= [(sin^{2} θ – cos^{2} θ) (sin^{2} θ + cos^{2} θ) + 1] cosec^{2} θ as [ sin^{2} θ + cos^{2} θ = 1]
= [(sin^{2} θ – cos^{2} θ) *1 + 1] cosec^{2} θ
= [sin^{2} θ – cos^{2} θ+1] cosec^{2} θ
= [(sin^{2} θ + (1 –cos^{2} θ)] cosec^{2} θ [ 1 – cos^{2} θ = sin^{2} θ]
= [sin^{2} θ + sin^{2} θ] cosec^{2} θ
= 2 sin2 θ . cosec^{2} θ
= 2 = RHS [∵ sin θ . cosec θ = 1]
Q4: Prove that: sec^{2} θ + cosec^{2} θ = sec^{2} θ · cosec^{2} θ
Sol: L.H.S. = sec^{2} θ + cosec^{2} θ
Q5:
Sol:
Q6: Given that α + β = 90°, show that:
Sol: ∵ α + β =90°
∵ β = (90 – a)
Q7:
Sol:
using a^{3}+b^{3}=(a+b)(a^{2}+b^{2}ab) and a^{3}b^{3}=(ab)(a^{2}+b^{2}+ab) in numerator of these terms
and also sin^{2}θ +cos2θ =1
Q8:
Sol:
using tanθ=sinθ/cosθ and then taking LCM
Q9:
Sol:
Q10: Prove that: sin^{6}θ + cos^{6}θ + 3sin^{2}θ cos^{2}θ = 1.
Sol: ∵ sin^{2 }θ + cos^{2}θ = 1
∵ (sin^{2} θ + cos^{2} θ)3 = (1)3 = 1
⇒ (sin^{2} θ)^{ 3} + (cos^{2} θ)^{3} + 3 sin^{2} θ . cos^{2} θ (sin^{2} θ + cos^{2} θ) = 1
⇒ sin^{6} θ + cos^{6} θ + 3 sin^{2} θ . cos^{2}θ (1) = 1
⇒ sin^{6} θ + cos^{6} θ + 3 sin^{2} θ . cos^{2} θ = 1
Q11: Prove that: a^{2} + b^{2} = x^{2} + y^{2} when a cos θ − b sin θ = x and a sin θ + b cos θ =y.
Sol:
Q12:
Sol:
Q13:
Sol:
Q14:
Sol:
Q15:
Sol:
Q16: For an acute angle θ, show that: (sin θ − cosec θ) (cos θ − sec θ)
Sol:
Q17:
Sol:
Q18:
Sol:
Q19:
Sol:
Q20: Without using trigonometric tables evaluate:
Sol:
Q21: If tan (A + B) = √3 and tan (A − B) = 1, 0° < A + B < 90°; A > B, then find A and B.
Sol: We have: tan (A + B) = √3 (Given)
tan 60° = √3 (From the table)
⇒ A + B = 60° ...(1)
Also,tan (A − B)= 1 [Given]
and cosec 60° = 2 and cos 90° = 0
⇒ A − B = 45 ...(2)
Adding (1) and (2),
2A = 60° + 45° = 105°
⇒
From (2), 52.5° − B = 45°
⇒ B = 52.5° − 45° = 7.5°
Thus, A = 52.5° and B = 7.5°.
Q22: If tan (2A) = cot (A − 21°), where 2A is an acute angle, then find the value of A.
Sol: We have: tan (2A) = cot (A − 21°)
∵ cot (90°− θ) = tan θ
∴ cot (90°− 2A) = tan 2A
⇒ cot (90°− 2A) = cot (A − 21)°
⇒ 90 − 2A = A − 21°
⇒ − 2A − A = − 21°− 90°
⇒ − 3A = − 111°
⇒
Q23: If sin 3A = cos (A − 10°), then find the value of A, where 3A is an acute angle.
Sol: We have:
sin 3A = cos (A − 10°)
∵ cos (90° − θ) = sin θ
∴ cos (90° − 3A) = sin 3A
⇒ 90° − 3A = A − 10°
⇒ −3A − A = −10° − 90°
⇒ −4A = −100°
⇒
Q24: If sec 2A = cosec (A − 27°), then find the value of A, where 2A is an acute angle.
Sol: We have:
sec 2A = cosec (A − 27°) ...(1)
∵ sec θ = cosec (90° − θ)
∴ sec 2A = cosec (90° − 2A) ...(2)
From (1) and (2), we get
A − 27° = 90° − 2A
⇒ A + 2A = 90 + 27° = 117°
⇒ 3A = 117°
⇒
Q25: Simplify:
+ sin θ cos θ
Sol: We have:
Q26:
Sol:
Q27:
Sol:
= 1 = R.H.S.
Q28: Without using trigonometrical tables, evaluate:
Sol:
[∵ sin (90° − θ) = cos θ, cos (90° − θ) = sin θ, cosec (90° − θ) = sec θ, and tan (90° − θ) = cot θ]
Q29: Using Geometry, find the value of sin 60°.
Sol: Let us consider an equilateral ΔABC and draw AD ⊥ BC.
Since, each angle of an equilateral triangle = 60°
∴∠A = ∠B = ∠C = 60°
Let AB = BC = AC = 2a
In ΔABD and ΔACD, we have:
AB = AC [Given]
∠ADB = ∠ADC = 90° [Construction]
AD = AD [Construction]
⇒ ΔABD ≅ ΔACD
⇒ BD = CD
Now, using Pythagoras theorem, in right ΔABD,
AD^{2} = AB^{2} − BD^{2}
= (2a)^{2} − a^{2}
= 4a^{2} − a^{2}
= 3a^{2}
⇒
∴
Thus,
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