Question

if tanc=tana+tanb/1+tanatanb then show that sin2c=sin2a+sin2b/1+sin2asin2b​

Answer 1

Given :    tanc= (tana+tanb)/(1+tanatanb )

To find : Show that sin2c=sin2a+sin2b/1+sin2asin2b​

Soluion:

Using Sin2x  = 2Tanx /(1 + tan²x)

LHS =  

sin2c  = 2Tanc /(1 + tan²c)

using tanc= (tana+tanb)/(1+tanatanb )

= 2(tana+tanb)/(1+tanatanb ) / ( 1  + { (tana+tanb)/(1+tanatanb )}²)

=   2(tana+tanb)(1+tanatanb) /( (1+tanatanb)² + (tana+tanb)²)

=  2(tana+  tan²atanb + tanb + tanatan²b ) /( 1 + tan²atan²b + 2TanaTanb + tan²a + tan²b + 2TanaTanb)

=  (2tana+  2tanb + 2tan²atanb  + 2tanatan²b ) /( 1 +  tan²a + tan²b  + tan²atan²b + 4TanaTanb )

RHS =  ( sin2a+sin2b)/(1+sin2asin2b​)

=  (2Tana /(1 + tan²a)  + 2Tanb /(1 + tan²b) )/(   1  + (2Tana /(1 + tan²a)) ( 2Tanb /(1 + tan²b))

= (  2Tana(1 + tan²b) +  2Tanb (1 + tan²a)  ) /( (1 + tan²a) (1 + tan²b) + 2Tana2Tanb)

=    (2Tana +  2Tanatan²b   +  2Tanb + 2tan²atanb) /( 1  +  tan²a + tan²b + tan²atan²b + 4TanaTanb )

=  (2tana+  2tanb + 2tan²atanb  + 2tanatan²b ) /( 1 +  tan²a + tan²b  + tan²atan²b + 4TanaTanb )

LHS = RHS

QED

hence proved

sin2c=sin2a+sin2b/1+sin2asin2b​

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