The formula of the sum of two angles of Tan Tan (PAI / 2 + a) = - COTA Tan (PAI / 2 - a) = COTA Why can these two formulas only be derived from tanq = SINQ / cosq? When we use the formula of the sum of two angles of Tan, Tan (a + b) = (Tana + tanb) / (1-tana * tanb), we can not deduce the result of this formula because there is Pie / 2; How can you explain that Tan (PAI / 2 + a) must be = - COTA? I only deduce this result from tanq = SINQ / cosq, but the result deduced from the formula of sum of two angles of Tan does not exist;

The formula of the sum of two angles of Tan Tan (PAI / 2 + a) = - COTA Tan (PAI / 2 - a) = COTA Why can these two formulas only be derived from tanq = SINQ / cosq? When we use the formula of the sum of two angles of Tan, Tan (a + b) = (Tana + tanb) / (1-tana * tanb), we can not deduce the result of this formula because there is Pie / 2; How can you explain that Tan (PAI / 2 + a) must be = - COTA? I only deduce this result from tanq = SINQ / cosq, but the result deduced from the formula of sum of two angles of Tan does not exist;

This formula doesn't need to be pushed, just remember that the trigonometric function values of the four quadrants are positive and negative, remember two sentences, odd variable and even constant, and look at the quadrant by sign, for example: for Tan (Pie / 2 * k + a) =? First look at K, odd or even, odd will become cot, and even will not change, and then take a as the angle of the first quadrant, and look at (Pie / 2 * k + a) after rotation

How to deduce the sum angle formula of Tan

tan(A+B)=sin(A+B)/cos(A+B)=(sinαcosβ+cosαsinβ)/(cosαcosβ-sinαsinβ)
At the same time divided by cosacosb
=(tanA+tanB)/(1-tanAtanB)

What is the formula of Tan (a + β)

tan(a+b)=sin（a+b）/cos（a+b）
=[sinacosb + cosasinb] / [cosacosb sinasinb] (divide cosacosb both above and below the fractional line) - the result is as follows
=(tana+tanb)/(1-tanatanb)

Tan (a + b) =? Formula

tan(A+B)=tanA+tanB/1-tanA*tanB
tan(A-B)=tanA-tanB/1+tanA*tanB