Solve the equation 2cot (x) = 1 + Tan (x) at 0

Solve the equation 2cot (x) = 1 + Tan (x) at 0

2/tanx=1+tanx
2=tanx+tan^2x (tanx≠0)
tan^2+tanx-2=0
(tanx+2)(tanx-1)=0
TaNx = - 2 or 1
So x = 2 π - arctan2 or π - arctan2 or π / 4 or 5 π / 4

Simplification
[sin(180+x)-tan(-x)+tan(-360-x)]/[tan(x+180)+cos(-x)+cos(-x-180)]

[sin(180+x)-tan(-x)+tan(-360-x)]/[tan(x+180)+cos(-x)+cos(-x-180)] =[-sinx+tanx-tan(360+x)]/[tanx+cosx+cos(180+x)]=[-sinx+tanx-tanx]/[tanx+cosx-cosx]=-sinx/tanx=-cosx

If f (TaNx) = sinxcosx, then the value of F (- 1) is?

Let x = - pi / 4, the result is f (- 1) = - 1 / 2