√12
= √4 * <span>√3
= 2</span><span>√3
------------------</span>
solution:
Z1 = 5(cos25˚+isin25˚)
Z2 = 2(cos80˚+isin80˚)
Z1.Z2 = 5(cos25˚+isin25˚). 2(cos80˚+isin80˚)
Z1.Z2 = 10{(cos25˚cos80˚ + isin25˚cos80˚+i^2sin25˚sin80˚) }
Z1.Z2 =10{(cos25˚cos80˚- sin25˚sin80˚+ i(cos25˚sin80˚+sin25˚cos80˚))}
(i^2 = -1)
Cos(A+B) = cosAcosB – sinAsinB
Sin(A+B) = sinAcosB + cosAsinB
Z1.Z2 = 10(cos(25˚+80˚) +isin(25˚+80˚)
Z1.Z2 = 10(cos105˚+ isin105˚)
Step-by-step explanation:
3x - 9 +2x + 9 = 2x - 2
5x = 2x - 2
5x - 2x = - 2
3x = - 2
x= - 2/ 3
Answer:
x = 5
Step-by-step explanation:
Using Pythagoras' identity in the right triangle.
The square on the hypotenuse is equal to the sum of the squares on the other 2 sides, that is
x² + 12² = 13², so
x² + 144 = 169 ( subtract 144 from both sides )
x² = 25 ( take the square root of both sides )
x = 
= 5
Answer:
(Sin A + Cos A)/Sin A. Cos A
Step-by-step explanation:
As we know
Sec A = 1/Cos A
and Cosec A = 1/Sin A
Given Equation
Sec A + Cosec A
Substituting the given values, we get -
1/cos A + 1/Sin A
(Sin A + Cos A)/Sin A. Cos A
