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If $\cot θ + \tan θ = x and \sec θ - \cos θ = y,$ then

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a

xsin⁡θ⋅cos⁡θ=1

b

sin2⁡θ=ycos⁡θ

c

x2y1/3+xy21/3=1

d

x2y2/3−xy22/3=1

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detailed solution

Correct option is A

cot⁡θ+tan⁡θ=xor cos⁡θsin⁡θ+sin⁡θcos⁡θ=xor 1=xsin⁡θcos⁡θ Now sec⁡θ−cos⁡θ=y1cos⁡θ−cos⁡θ=yor sin2⁡θ=ycos⁡θnow x2y=1sin2⁡θcos2⁡θ⋅sin2⁡θcos⁡θ=1cos3⁡θand xy2=sin3⁡θcos3⁡θ∴x2y2/3−xy22/3=1cos2⁡θ−sin2⁡θcos2⁡θ=1

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