Question

Simplify: cot(β − γ) cot(γ − α) + cot(γ − α) cot(α − β) + cot(α − β) cot(β − γ) ​

Answer 1

SOLUTION

TO SIMPLIFY

cot(β − γ) cot(γ − α) + cot(γ − α) cot(α − β) + cot(α − β) cot(β − γ)

EVALUATION

Let x = (α − β) , y = (β − γ) , z = (γ − α)

Then

x + y + z

= (α − β) + (β − γ) + (γ − α)

= 0

Thus x + y = - z

$\displaystyle\sf{ \implies \: \cot(x + y) = \cot( - z)}$

$\displaystyle\sf{ \implies \: \cot(x + y) = - \cot z}$

$\displaystyle\sf{ \implies \: \frac{ \cot x \cot y - 1}{ \cot x + \cot y} = - \cot z}$

$\displaystyle\sf{ \implies \: \cot x \cot y - 1 = - \cot x \cot z - \cot y \cot z}$

$\displaystyle\sf{ \implies \: \cot x \cot y + \cot x \cot z + \cot y \cot z = 1}$

⇒ cot(β − γ) cot(γ − α) + cot(γ − α) cot(α − β) + cot(α − β) cot(β − γ) = 1

FINAL ANSWER

cot(β − γ) cot(γ − α) + cot(γ − α) cot(α − β) + cot(α − β) cot(β − γ) = 1

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