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Questions  

If sinαsinβcosαcosβ+1=0 then 1+cotαtanβ=

a
-1
b
1
c
0
d
2

detailed solution

Correct option is C

We have sinαsinβ-cosαcosβ=-1⇒cos(α+β)=1⇒α+β=0,2π,4π ……Now 1+cotαtanβ=1+cosαsinα·sinβcosβ                                =sin(α+β)sinαsinβ                                 =0       ∵α+β=2nπ

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