In △ ABC, the following expression is constant: A. sin (a + b) + sinc B. cos (B + C) - cosa C. Tan (a + b) / 2 times Tanc / 2 D. Cos (B + C) / 2 times Tana / 2

In △ ABC, the following expression is constant: A. sin (a + b) + sinc B. cos (B + C) - cosa C. Tan (a + b) / 2 times Tanc / 2 D. Cos (B + C) / 2 times Tana / 2

A+B=180-C
So tan (a + b) / 2 times Tan C / 2
=Tan (180-c) / 2 times Tan C / 2
=Tan (90-c / 2) multiplied by Tan C / 2
=COTC / 2 times Tanc / 2
=1
Choose C

In the triangle ABC, cos (a + b) > 0, sinc = 1 / 3, Tanc = =?

cos(a+b)>0
a+b90
sinc=1/3
cosc=-2√2/3
tanc=-√2/4

In the triangle ABC, if cos (a + b) > 0, sinc = 1 / 3, then Tanc=

The specific solving process is as follows:
Cos (a + b) > 0 description (a + b)

The lengths of the opposite sides of the three interior angles a B C of the triangle ABC are a B C and a = 2 √ 3, Tan (a + b) / 2 + Tanc / 2 = 4, SINB * sinc = cos ^ 2A / 2
Finding a, B and B, C

(1)
tan[（A+B）/2]+tanC/2=4,
And Tan (a + b) = Tan (π - C)
So tan [(a + b) / 2] + Tan C / 2
=tan[π/2-C/2]+tanC/2
=cot(C/2)+tan(C/2)
=cos(C/2)/sin(C/2)+sin(C/2)/cos(C/2)
=[sin(C/2)^2+cos(C/2)^2]/(sinC/2)(cosC/2)=2/sinC
SO 2 / sinc = 4, that is sinc = 1 / 2
So C = π / 6 or 5 π / 6