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1.

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

Answer»

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 (A + B) = \(\frac{tanA+ tanB}{1-tanA\,tanB}\)

\(\frac{tan2A+ tan2B}{1-tan2A\,tan2B}\) = - tan 2C

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
Discussion
2.

If A + B + C = π, prove that sin 2A + sin 2B – sin 2C = 4cos A cos B sin C

Answer»

sin 2A + sin 2B – sin 2C 

= 2 sin (B + C) cos A + 2 sin (A + C) cos B - 2 sin (A + B) cos C 

Using formula, sin (A + B) = sin A cos B + cos A sin B 

= sin 2A + sin 2B - sin 2C 

Using formula 

sin2A = 2sinAcosA 

= 2sinAcosA + 2sinBcosB - 2sinCcosC 

Since A + B + C = π 

→ B + C = 180 - A

And sin(π – A) = sinA 

= 2sin(B + C)cos A + 2sin(A + C)cosB - 2sin(A + B)cosC 

= 2 ( sin B cos C + cos B sin C ) cos A + 2(sinAcosC + cosAsinC)cosB - 2(sinAcosB + cosAsinB )cosC 

= 2cosAsinBcosC + 2cosAcosBsinC + 2sinAcosBcosC + 2cosAcosBsinC– 2sinAcosBcosC – 2cosAsinBcosC 

= 2cosAcosBsinC + 2cosAcosBsinC 

= 4cosAcosBsinC 

= R.H.S
Discussion