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Question
In ΔABC, A + B + C = π show that
tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C
Solution
In ΔABC,
A + B + C = π
∴ 2A + 2B + 2C = 2π
∴ 2A + 2B = 2π – 2C
∴ tan (2A + 2B) = tan (2π – 2C)
∴ `(tan2"A" + tan2"B")/(1 - tan2"A"*tan2"B")` = – tan 2C
∴ tan 2A + tan 2B = − tan 2C . (1 − tan 2A . tan 2B)
∴ tan 2A + tan 2B = – tan 2C + tan 2A· tan 2B· tan 2C
∴ tan 2A + tan 2B + tan 2C = tan 2A · tan 2B· tan 2C
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