`[3(cos40+isin40)]*[4(cos80+isin80)]=` (A) ` -1+sqrt3i ` (B) ` 4[-1+sqrt3i]` (C) ` 2[-1+sqrt3i] ` (D) ` 6[-1+sqrt3i] `

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[3(cos40+i sin40)]*[4(cos80+i sin80)]=(A)-1+sqrt(3)i(B)4[-1+sqrt(3)i](C)2[-1+sqrt(3)i](D)6[-1+sqrt(3)i]

If : sin 40^(@)-cos 70^(@)= k * cos 80^(@), "then" : k = A) sqrt1 B) sqrt2 C) sqrt3 D) sqrt4

What is (sqrt3+i)//(1+sqrt3i) equal to ?

(1) /( cos 80^@ ) - ( sqrt(3))/( sin 80 ^@) =

log((1+i sqrt(3))/(1-i sqrt(3)))

The modulus of (sqrt3 + i)/((1 + i) (1 + sqrt3i)) is

1/(cos80^(@))-sqrt(3)/(sin80^(@)) is equal to a) sqrt(2) b) sqrt(3) c)2 d)4

(sqrt(3)+i)/((1+i)(1+sqrt(3)i))|=