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If tanθ = a sinϕ/1-acosϕ and tanϕ = bsinθ/1-bcosθ then a/b= (a) sinθ/1-cosθ (b) sinθ/1-cosϕ (c) sinϕ/sinθ (d) sinθ/sinϕ

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If tanθ = a sinϕ/1-acosϕ and tanϕ = bsinθ/1-bcosθ then a/b= 

(a) sinθ/1-cosθ

(b) sinθ/1-cosϕ

(c) sinϕ/sinθ

(d) sinθ/sinϕ

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The correct option is (d).

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