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If f(x)=sqrt("cosec"^2x-2sinx cosx - 1/(...

If `f(x)=sqrt("cosec"^2x-2sinx cosx - 1/(tan^2x)) x in ((7pi)/4,2pi)`, then `f.((11pi)/6)=`

A

`(1-sqrt3)/2`

B

`(sqrt3-1)/2`

C

`(1+sqrt3)/2`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
A
`f(x)=sqrt("cosec"^2x-cot^2x-2sin x cos x)`
`=sqrt((cosx-sinx)^2)`

f(x)=cos x -sin x
`rArr` f.(x)=-sin x - cos x
`f.((11pi)/6)=-[sin (11pi)/6 + cos(11pi)/6]`
`rArr f.((11pi)/6)=-[(-1)/2+sqrt3/2]=(1-sqrt3)/2`
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