Question

Hello please help me to solve these

© if A + B + C = 180 ° and SinA + SinBSinC = 0 then prof that 2 TanB + TanC = 0

Answer 1

Hey there !!!!

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A+B+C=180°

A= 180°-(B+C)

Applying "sin" on both sides

sinA =sin(180-(B+C))--- Equation1

We know that sin(180-∅) = sin∅

sin(180-(B+C) ) =sin(B+C)

So, equation1 changes to

sinA= sin(B+C)

According to question sinA+sinBcosC=0

sinA=-sinBcosC

sin(B+C) = -sinBcosC=sinA

Now

2TanB+TanC

= 2sinB/cosB + sinC/cosC

= (2sinBcosC+sinCcosB)/cosBcosC

=sinBcosC+sinBcosC+sinCcosB/cosBcosC

But sinBcosC+sinCcosB=sin(B+C)

=sinBcosC+sin(B+C)/cosBcosC

= sinBcosC+sinA/cosBcosC

=sinBcosC-(sinBcosC)/cosBcosC

= 0

LHS=RHS

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Hope this helped you.......

Answer 2

Hi ,

A + B + C = 180

B + C = 180 - C

take sin both sides,

sin( B + C ) = Sin ( 180 - A )

= Sin A ------( 1 )

SinA + SinBSinC = 0 ----( 2 )

according to the problem given,

LHS = 2 TanB+ Tan C

= Tan B+ Tan B + Tan C

= SinB / CosB + SinB / CosB + SinC/CosC

= (sinBCosC+SinBCosC+SinCCosB)/cosBcosC

= [sinBcosC+(sinBcosC+sinCcosB)]/cosBcosC

=[sinAcosC+sin(B+C)]/(cosBcosC

[since sinBcosC+sinCcosB = sin(B+C) ]

= [ sinAcosC+SinA ] /( cosBcosC) [ from ( 1 )]

= 0 / cosBcosC [ from ( 2 ) ]

= 0

= RHS

Hence proved.

I hope this helps you.

:)