Question

Please someone help me to prove this. ​

Answer 1

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)

Answer 2

There is an approach that applies 'inverse trigonometric functions.'

Let's say that A + B + C = π.

Given: A + B + C = π,

We can conclude: A/2 + B/2 + C/2 = π/2

Now let's say that A/2 = tan⁻¹x, B/2 = tan⁻¹y, and C/2 = tan⁻¹z.

We can conclude: tan⁻¹x + tan⁻¹y + tan⁻¹z = π/2

If we simplify the equation 'tan⁻¹x + tan⁻¹y + tan⁻¹z = π/2' we should receive the equation 'tan⁻¹x + tan⁻¹y = cot⁻¹z.' This is as 'tan⁻¹z + cot⁻¹z = π/2.' Further simplification would be as follows,

$\mathrm{tan^{-1}\left(\frac{x+y}{1-xy}\right)=\:cot^{-1}z,}\\\mathrm{cot^{-1}\left(\frac{1-xy}{x+y}\right)=\:cot^{-1}z,}\\\mathrm{\left(\frac{1-xy}{x+y}\right)=\:z,}\\\mathrm{\left(\frac{1}{xyz}\right)=\frac{1}{z}+\frac{1}{x}+\frac{1}{y}\:}$

Therefore, if we substitute our assigned values back, cotA/2 $*$ cot B/2 $*$ cot C/2 = cot A/2 + cot B/2 + cot C/2. Hence proved.