Question

In triangle ABC, prove that cot( A/2)+cot( B/2)+cot(C/2)=(a+b+c/b+c-a)cot(A/2)​

Answer 1

Answer:

hope this may help you

Answer 2

Answer:

Step-by-step explanation:

Given: A ΔABC having its side as AB, BC, CA and having angles as ∠A, ∠B and ∠C.

A/2 + B/2 = pi/2 - C/2

Cot(A/2 + B/2) = Cot(pi/2 - C/2)

A/2 + B/2 = π/2 − C/2

cot(A/2 + B/2) = cot(π/2 − C/2) = tanC/2 = 1/cotC/2

Apply cot(x + y) = cotxcoty − 1/cotx + coty

= tanC/2 = 1/Cot(c/2)

Apply cot(x + y) = (Cotxcoty - 1)/(cotx + coty)

cot A/2 + cot B/2 + cot c/2 = a + b + c/b + c - a cot (B/2) = a + b + c/ a + b - c cot(C/2)

Hence, cot( A/2)+cot( B/2)+cot(C/2)=(a+b+c/b+c-a)cot(A/2)​

Learn all these formula's hope it will help you .

If none of the angles A, B and (A ± B) is an odd multiple of π/2, then

• tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]
• tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

If none of the angles A, B and (A ± B) is a multiple of π, then

• cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]
• cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

Some additional formulas for sum and product of angles:

• cos(A+B) cos(A–B)=cos2A–sin2B=cos2B–sin2A
• sin(A+B) sin(A–B) = sin2A–sin2B=cos2B–cos2A
• sinA+sinB = 2 sin (A+B)/2 cos (A-B)/2

Formulas for twice of the angles:

• sin2A = 2sinA cosA = [2tan A /(1+tan2A)]
• cos2A = cos2A–sin2A = 1–2sin2A = 2cos2A–1= [(1-tan2A)/(1+tan2A)]
• tan 2A = (2 tan A)/(1-tan2A)

Formulas for thrice of the angles:

• sin3A = 3sinA – 4sin3A
• cos3A = 4cos3A – 3cosA
• tan3A = [3tanA–tan3A]/[1−3tan2A]

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