`(sin 70^@ + cos 40^@)/(cos 70^@ + sin 40^@)`

Explanation

`(sin70^@ +cos40^@)/(cos70^@+sin40^@)`

We know, `sintheta = cos(90-theta)`. So, our expression becomes

`(cos20^@+cos40^@)/(sin20^@+sin40^@)`

As ,`cosA+cosB = 2cos((A+B)/2)cos((A-B)/2)`

and `sinA+sinB = 2sin((A+B)/2)cos((A-B)/2)`

Our expression becomes,

`=(2cos((20^@+40^@)/2)cos((20^@-40^@)/2))/(2sin((20^@+40^@)/2)cos((20^@-40^@)/2))`

`=cot30^@ = sqrt3`