CDS 2026 · PYQ · Trigonometric Identities · hard
p + q cotθ = 3cosecθ and q − p cotθ = 2cosecθ. What is tanθ equal to?
- A.(2q+3p)/(3q−2p)✓ Correct
- B.(2q−3p)/(3q+2p)
- C.(3q+2p)/(2q−3p)
- D.(3q−2p)/(2q+3p)
Multiply first equation by p: p² + pq cotθ = 3p cosecθ. Multiply second by q: q² − pq cotθ = 2q cosecθ. Adding: p² + q² = (3p + 2q)cosecθ, so cosecθ = (p² + q²)/(3p + 2q) = 13/(3p + 2q). Multiply first by q and second by p: pq + q²cotθ = 3q cosecθ; pq − p²cotθ = 2p cosecθ. Subtracting: (p² + q²)cotθ = (3q − 2p)cosecθ, so cotθ = (3q − 2p)/(3p + 2q) × … using cosecθ value: cotθ = (3q − 2p)cosecθ/13 = (3q − 2p)/(3p + 2q). Hence tanθ = (3p + 2q)/(3q − 2p) = (2q + 3p)/(3q − 2p).
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