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

*print*Print*list*Cite

### 1 Answer

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)` **