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Question

# If cotθ+tanθ=x and secθ−cosθ=y, then

A
sinθcosθ=1x
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B
sinθtanθ=y
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C
(x2y)23(xy2)23=1
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D
(x2y)13(xy2)13=1
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Solution

## The correct options are A sinθcosθ=1x B sinθtanθ=y C (x2y)23−(xy2)23=1Since cotθ+tanθ=x andsecθ−cosθ=yoption (A):∴1x=1cotθ+tanθ=sinθcosθcos2θ+sin2θ=sinθcosθ1∴sinθcosθ=1xoption (B):since, y=secθ−cosθ=1−cos2θcosθ=sin2θcosθsinθtanθHence, sinθtanθ=yoption (C):since, x2y=1sin2θcos2θ×sin2θcosθ=1cos3θ⇒(1x2y)23=cos2θ⇒(x2y)23=sec2θ....(i)and xy2=1sinθcosθ×sin4θcos2θ=tan3θ⇒(xy2)23=tan2θ....(ii)From equation (i) and (ii),(x2y)23−(xy2)23=1

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