Question

In triangle ABC,prove that; tan B-C/2=b-c/b+c.cotA/2

Answer 1

Answer:

tan(B−C)2=b−cb+ccotA2tan(C−A)2=c−ac+acotB2tan(A−B)2=a−ba+bcotC2

Explanation:

From sine law, we have, asinA=bsinB=csinC=k

∴a=ksinA,b=ksinB,c=ksinC

∴b−cb+c=k(sinB−sinC)k(sinB+sinC)

=sinB−sinCsinB+sinC

=2cos(B+C2)sin(B−C2)2sin(B+C2)cos(B−C2)

=cot(B+C2)tan(B−C2)

Here, A+B+C=180

So, B+C=180−A

∴b−cb+c=cot(180−A2)tan(B−C2)=cot(90−A2)tan(B−C2)

b−cb+c=tan(A2)tan(B−C2)

tan(B−C2)=b−cb+ccot(A2)

Similarly, we can show that,

tan(C−A2)=c−ac+acot(B2)

tan(A−B2)=a−ba+bcot(C2)