Show that tg40 +tg20 = square root of3 sec10^2

You need to remember the formula `tan a + tan b =
sin(a+b)/(cos a*cos b)`

`` `tan20 + tan 40 =
sin(20+40)/(cos 20*cos 40) = sin 60/(cos 20*cos 40)`

To
calculate the product found at denominator, you need the help of
formula:

`2cos a* cos b = cos(a+b) + cos(a-b)
`

`2cos 20*cos 40 = cos(20+40) +
cos(20-40)`

`2cos 20*cos 40 = cos 60 +
cos(-20)`

The function cosine is even => cos(-20) =
cos 20

`2cos 20*cos 40 =1/2 +
cos(20)`

`tan20 + tan 40 = 2((sqrt3)/2)/(1/2 +
cos(20))`

`tan20 + tan 40 = 2((sqrt3))/(1+
2cos(20))`

`cos 20 = cos2*10 = 2cos^2 10 -
1`

`tan20 + tan 40 = 2((sqrt3))/(1+2cos^2 10 -
1)`

Remove opposite
terms=>

=> `tan20 + tan 40 =
2((sqrt3))/(2cos^2 10) = (sqrt3)/(cos^2 10)`

Remember that
`sec 10 = 1/ (cos 10)`

`tan20 + tan 40 = (sqrt3)*sec^2
(10)`

ANSWER: `tan20 + tan 40 = (sqrt3)*sec^2
(10)`