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# Prove that in any triangletanB-C = b-c cotA/2     2       b+c

Grade:11

## 3 Answers

BALAJI ANDALAMALA
askIITians Faculty 78 Points
5 years ago
Rp mahaveer
13 Points
3 years ago
Consider b-c/b+c=2RSINB-2RSINC/2RSINB+2RSINC = 2R (SINB-SINC)/(SINB+SINC) =2COS(B+C/2)×SIN(B-C/2)/2SIN(B+C/2)×COS(B-C/2) =COT (B+C/2)×TAN(B-C/2)=COT (90°-A/2)×TAN (B-C/2)=TANA/2×TAN(B-C/2)TAN(B-C/2)=B-C/B+C×COT(A/2)
Kushagra Madhukar
askIITians Faculty 629 Points
10 months ago
Dear student,
Please find the solution to your problem.

Consider b-c/b+c
= 2RsinB – 2RsinC/2RsinB + 2RsinC
= 2R(sinB – sinC)/(sinB + sinC)
= 2cos(B + C/2) × sin(B – C/2)/2sin(B + C/2) × cos(B – C/2) = cot(B + C/2) × tan(B – C/2)
= cot(90° – A/2) × tan(B – C/2)
= tanA/2 × tan(B – C/2)
Hence, tan(B – C/2) = b-c/b+c × cot(A/2)

Thanks and regards,
Kushagra

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