Q. 12

If A + B + C = π, prove thattan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

= tan 2A + tan 2B + tan 2C

Since A + B + C = π

A + B = π – C

2A + 2B = 2π – 2C

Tan (2A+2B) = tan (2π – 2C)

Since tan (2π – C) = -tan C

Tan (2A + 2B) = -tan 2C

Now using formula,

Tan 2A + tan 2B = -tan 2C + tan 2C tan 2B tan 2A

Tan 2A + tan 2B + tan 2C = tan 2A tan 2B tan 2C

= R.H.S

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