Grade 11Trigonometry
Tan(x-y)+tan(y-z)+tan(z-x):::tan(x-y)tan(y-z)tan(z-x)
1 Answer
Arun
tan(z-x) = tan((y-x)-(y-z))
=> tan(z-x) = (tan(y-x) - tan(y-z))/(1+tan(y-x).tan(y-z))
by given formula.
Multiply both sides by (1+tan(y-x).tan(y-z)):
tan(z-x).(1+tan(y-x).tan(y-z)) = tan(y-x)-tan(y-z)
=> -tan(y-x) + tan(y-z) + tan(z-x) = -tan(y-x).tan(y-z).tan(z-x)
=> tan(x-y) + tan(y-z) + tan(z-x) = tan(x-y).tan(y-z).tan(z-x),
since tan(x-y) = -tan(y-x).
=> tan(z-x) = (tan(y-x) - tan(y-z))/(1+tan(y-x).tan(y-z))
by given formula.
Multiply both sides by (1+tan(y-x).tan(y-z)):
tan(z-x).(1+tan(y-x).tan(y-z)) = tan(y-x)-tan(y-z)
=> -tan(y-x) + tan(y-z) + tan(z-x) = -tan(y-x).tan(y-z).tan(z-x)
=> tan(x-y) + tan(y-z) + tan(z-x) = tan(x-y).tan(y-z).tan(z-x),
since tan(x-y) = -tan(y-x).
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