Trigonometry Problems and Solutions - 2
Trigonometry Problems and Solutions -2
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| Trigonometry Kumar-Mathematics |
(P) If cos2θ =sin4θ, where 2θ and 4θ are acute angles, find the value of θ.
Solution : cos2θ =sin4θ
⇒ cos2θ =cos (90°–4θ) {∵ cos(90°–θ)= sinθ}
⇒ 2θ = 90°–4θ
⇒ 2θ +4θ= 90°
⇒ 6θ = 90°
⇒ θ = 15°
(P) sin3θ = cos(θ –6°), where 3θ and (θ–6°) are acute angles, find the value of θ
Ans
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sin3θ = cos(θ –6°) ⇒ cos(90° –3θ) = cos(θ –6°) { ∵ sinθ = cos(90°–θ)}
⇒ 90°–3θ = θ – 6°
⇒ 90°+6°= θ +3θ
⇒ 96° = 4θ
⇒ θ = 24°
Solution : sec2A = cosec(A–42°)
⇒cosec(90°–2A) = cosec(A–42°) {∵ cosec(90°–A)= sec(A)}
⇒ 90°–2A = A – 42°
⇒ 90°+42°=A+2A
⇒ 132°=3A
⇒ A=132°/3
= 44°
(P) If sec4A = cosec(A–20°), where 4A is an acute angle, find the value of A.
Solution : sec4A = cosec(A–20°)
⇒cosec(90°–4A) = cosec(A–20°) {∵ cosec(90°–A)= sec(A)}
⇒ 90°–4A = A – 20°
⇒ 90°+20°=A+4A
⇒ 110°=5A
⇒ A=110°/5
= 22°
Formula
✦ tanAtanB=1 if A+B =90°
[∵ A+B =90° ⇒ B=90°– A
Let tanAtanB
⇒ tanAtan(90°– A)
⇒ tanAcotA =1
∴ tanA.tanB=1 if A+B =90°]
Choose one correct answer
(P) The value of tan1°tan2°tan3°………..tan89° is
1 –1 0 None of these
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tan1°tan2°tan3°………..tan89°tan1°tan2°tan3°………..tan89°
⇒tan1°tan89°tan2°tan88°………..tan44°tan46°tan45°
⇒ (1)(1)........(1)(1) [∵ tanAtanB=1 if A+B=90°]
=1
–1 0 1 None of these
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tan10°tan15°tan75°tan80°⇒ tan10°tan80°tan15°tan75°
⇒ (1) (1) { ∵tanAtanB=1 if A+B=90°}
=1
1 0 –1 None of these
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cos1°cos2°cos3°…….cos180°⇒ cos1°……..cos90°…….cos180°
⇒ cos1°……..(0)…….cos180°
= 0
2 – 2 – 1/ 2 1/2
check answer
tan²45° – cos²30°= x sin45°cos45° ⇒ (1)² – (√3/2)² = x (1/√2)(1/√2)
⇒ 1 – (3/4)=x/2
⇒1/4 = x/2
⇒x = 1/2
1 1/3 0 –1
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cos²17° – sin²73° ⇒ cos²17° – {sin(90° – 17°)}²
⇒ cos²17° – cos²17° {∵ sin(90°–θ) = cosθ}
= 0
