Please someone help me to prove this.

Step-by-step explanation:

This is known as the triple cotangent identity, and it is only valid if the three angles add up to π/2.

A/2 + B/2 + C/2 = π/2

Let's define the complement of each angle:

α = π/2 − A/2

β = π/2 − B/2

γ = π/2 − C/2

Add the complements together:

α + β + γ

= (π/2 − A/2) + (π/2 − B/2) + (π/2 − C/2)

= 3π/2 − (A/2 + B/2 + C/2)

= 3π/2 − π/2

= π

Therefore, the triple tangent identity applies.

tan α tan β tan γ = tan α + tan β + tan γ

Substitute:

tan(π/2 − A/2) tan(π/2 − B/2) tan(π/2 − C/2) = tan(π/2 − A/2) + tan(π/2 − B/2) + tan(π/2 − C/2)

Use reflection identity:

cot(A/2) cot(B/2) cot(C/2) = cot(A/2) + cot(B/2) + cot(C/2)

for b its (t) and c its (d)

step-by-step explanation:

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.