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If p = cotθ + tanθ and q = secθ – cosθ, then (p2q)2/3 – (q2p)2/3 is equal to
- a)0
- b)1
- c)2
- d)3
Correct answer is option 'B'. Can you explain this answer?
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If p = cotθ + tanθ and q = secθ – cosθ, ...
p = cotθ + tanθ = cosθ/sinθ + sinθ/cosθ
⇒ (cos2θ + sin2θ)/(sinθ cosθ)
⇒ 1/ (sinθ cosθ) [∵ cos2θ + sin2θ = 1]
⇒ p = 1/ (sinθ cosθ)
q = secθ – cosθ = 1/cosθ – cosθ [∵ 1/cosθ = secθ]
⇒ (1 – cos2θ) /cosθ
⇒ sin2θ/cosθ [∵ 1 – cos2θ = sin2θ]
⇒ q = sin2θ/cosθ
p2q = [1/(sinθ cosθ)]2 × sin2θ/cosθ = 1/cos3θ
⇒ p2q = sec3θ
pq2 = 1/(sinθ cosθ) × (sin2θ/cos θ)2 = sin3θ/cos3θ
⇒ pq2 = tan3 θ
∴ (p2q)2/3 – (q2p)2/3 = sec2θ – tan2θ = 1
⇒ (cos2θ + sin2θ)/(sinθ cosθ)
⇒ 1/ (sinθ cosθ) [∵ cos2θ + sin2θ = 1]
⇒ p = 1/ (sinθ cosθ)
q = secθ – cosθ = 1/cosθ – cosθ [∵ 1/cosθ = secθ]
⇒ (1 – cos2θ) /cosθ
⇒ sin2θ/cosθ [∵ 1 – cos2θ = sin2θ]
⇒ q = sin2θ/cosθ
p2q = [1/(sinθ cosθ)]2 × sin2θ/cosθ = 1/cos3θ
⇒ p2q = sec3θ
pq2 = 1/(sinθ cosθ) × (sin2θ/cos θ)2 = sin3θ/cos3θ
⇒ pq2 = tan3 θ
∴ (p2q)2/3 – (q2p)2/3 = sec2θ – tan2θ = 1
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If p = cotθ + tanθ and q = secθ – cosθ, then (p2q)2/3– (q2p)2/3is equal toa)0b)1c)2d)3Correct answer is option 'B'. Can you explain this answer?
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