Question

prove that tan(y - Z) + tan(z - x) + tan(x - y) = tan(x - y).tan (y-z) tan(z-x)​

Answer 1

Answer:

[tan(x - y) tan(y - z) tan(z - x) = tan(x - y) + tan(y - z) + tan(z - x).

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Concepts

We know that

$\tan( \alpha + \beta + \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) + \tan( \gamma ) - \tan( \alpha ) \tan( \beta ) \tan( \gamma ) }{1 - \tan( \alpha ) \tan( \beta ) - \tan( \beta ) \tan( \gamma ) - \tan( \gamma ) \tan( \alpha ) }$

This can also be proved using the formula

$\tan( \alpha + \beta ) = \frac{ \tan( \alpha ) \tan( \beta ) }{1 - \tan( \alpha ) \tan( \beta ) }$

In the above formula if

$\alpha + \beta + \gamma = 0 \: \: then$

$\tan( \alpha ) + \tan( \beta ) + \tan( \gamma ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma )$

Proof

If we take α = y-z

β = z-x

γ = x-y

Then α+β+γ = (y-z)+(z-x)+(x-y) = 0

Therefore, from the above discussion

$\tan(y - z) + \tan(z - x) + \tan(x - y) = \tan(y - z) \tan(z - x) \tan(x - y)$

Hence Proved.