Question

sin(A-B)(tanA+tanB)=sin(A+B)(tanA-tanB). Proof

Answer 1

Sin(A-B) (TanA+TanB) =Sin(A+B) (TanA-TanB)
Taking L. H. S. first
=(SinA-SinB) (SinA/CosA+SinB/CosB)
=Sin^2A/CosA-Sin^2B/CosB
Now taking R. H. S....
=(SinA+SinB) (SinA/CosA-SinB/CosB)
=Sin^2A/CosA-Sin^2B/CosB
Hence.. L. H. S=R. H. S. proved
I hope that u got that.. I have skipped some steps u should get them by solving it in ur copy

Answer 2

hello users ......

we have to prove that 
Sin(A-B)(tanA+tanB)=sin(A+B)(tanA-tanB)

solution ;-
formula's used:=
sin ( a-b) = sin a cos b - cos a sin b 
and 
sin ( a+ b) = sin a cos b + cos a sin b 
and 
tan x = sin x/ cos x
and
a²-b² = ( a+b) ( a-b)


now

taking LHS 
 Sin(A-B)(tanA+tanB)
= (sin A cos B - cos A sin B) ( sin A/ cos A + sin B / cos B ) 

=  (sin A cos B - cos A sin B)( sin A cos B + cos A sin B ) / cos A cos B

= {( sin A cos B)² - (cos A sin B)² } / cos A cos B 

now taking RHS 

sin(A+B)(tanA-tanB)
 =  (sin A cos B + cos A sin B )(  sin A/ cos A - sin B / cos B)
 
= (sin A cos B + cos A sin B ) ( sin A cos B - cos A sin B ) / cos A cos B 

=  {( sin A cos B)² - (cos A sin B)² } / cos A cos B 

hence ;

LHS = RHS   proved ;

❈❈hope it helps ❈❈