Question

In a triangle abc,tan a+b/2 cot a-b/2 is equal to

Answer 1

@campbell_st

Answer 2

is it

\[\tan(\frac{a + b}{2}) + \cos(\frac{a - b}{2})\]

Answer 3

nope cot a-b/2

Answer 4

\[\tan(\frac{A + B}{2}) + \cot(\frac{A - B}{2})\]

like this ?

Answer 5

tan (a+b)/2*cot(a-b)/2

Answer 6

\[\tan(A+B)/2*\cot(A+B)/2\]

Answer 7

sorry cot (a-b)/2

Answer 8

\[\tan(\frac{A + B}{2}) * \cot(\frac{A - B}{2})\]

Answer 9

\(\large A + B + C = \pi\)

Answer 10

c=pi-a+b

Answer 11

a+b=pi-c

Answer 12

\[\tan(\frac{A + B}{2}) * \cot(\frac{A - B}{2})\]

\[\tan(\frac{C-\pi}{2}) * \cot(\frac{A - B}{2})\]

\[\cot(\frac{C}{2}) * \cot(\frac{A - B}{2})\]

Answer 13

(a - b)/(a + b) = tan [(A-B)/2] / tan [(A+B)/2] (Law of Tangents)

Answer 14

Oh it equals (a-b)/(a+b) is it ? xD

Answer 15

yes

Answer 16

so a+b/a-b=cot(a-b)/2*tan(a+b)/2 correct