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If f(x)=sqrt("cosec"^2x-2sinx cosx - 1/(tan^2x)) x in ((7pi)/4,2pi), then f.((11pi)/6)= |
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Answer» `(1-sqrt3)/2` `=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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