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Q.

tanAtanB=xand cotBcotA=4, then cot(AB)=

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a

1xy

b

1x1y

c

1x+1y 

d

x+y

answer is A.

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Detailed Solution

y=cotBcotA=1tanB1tanA=xtanAtanB

tanAtanB=xy

Now, 

cot(AB)=1tan(AB)

=1+tanAtanBtanAtanB=1+xyx=1x+1y

 

 

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tan⁡A−tan⁡B=xand cot⁡B−cot⁡A=4, then cot⁡(A−B)=