Q.

cosx+cosy+cosα=0 and sinx+siny+sinα=0 then cotx+y2=

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a

sinα

b

cosα

c

tanα

d

cotα

answer is D.

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Detailed Solution

cosx+cosy=−cosαsinx+siny=−sinα⇒2cosx+y2cosx−y2=−cosα and 2sinx+y2cosx−y2=−sinαby dividing ∴cotx+y2=cotα
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