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Qus: If tanθ= (x sinϕ) / (1−xcosϕ) and, tan ϕ = (y sin θ)/ (1−y cos θ) then x/y =


A. sinθ / (1−cosϕ)
B. sinθ / (1−cosθ)
C. sinθ/sinϕ
D. sinϕ / sinθ


Solution
C. sinθ/sinϕ



Explanation

We have, tanθ = (x sinϕ)/ (1−xcosϕ)


⇒ (1−xcos ϕ) / (x sin ϕ) = 1/ tanθ ⇒ (1/ xsin ϕ) −cotϕ=cotθ


⇒ 1/ xsin ϕ= =cot θ+cot ϕ


and tan ϕ = y sinθ / (1−y cosθ) ⇒ (1−y cosθ)/ y sinθ = 1/ tan ϕ


⇒ (1/y sin θ) – cot θ = cotϕ⇒ (1/ y sin θ) =cot ϕ+cot θ


⇒ (1/y sin θ) = (1/ x sin ϕ) ⇒ x/y = sin θ/ sin ϕ

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