Question

Sin 2a= x sin 2b prove that tan(a+b)/tan(a-b) = x+1/x-1

Answer 1

Answer:

this question is actually very long so forgive me if you cant understand

Step-by-step explanation:

sin2A=2tanA/1+tan²A

sin2B=2tanB/1+tan²B

cross multiply in the given equation after substituting the values

2tanA*(1+tan²B)/1+tan²A(2tanB)=x

x=2tanA+2tan²BtanA/2tanB+2tan²AtanB

now find values of x+1 and x-1 separately

x+1=2tanA+2tan²BtanA+2tanB+tan²AtanB (I am not writing denominator in x+1and x-1 because it will be cancelled off anyway when we divide x+1/x-1)

x-1=2tanA+2tan²BtanA-2tanB-tan²AtanB

x+1/x-1=2(tan +tanB)+2(tan²AtanB+tan²BtanA)/2(tanA-tanB)+2(tan²AtanB-tan²BtanA)

=>tanA+tanB(1+tanAtanB)/tanA-tanB(1+tanAtanB)

now if you remember the formula the above would be equated to

tan(A+B)/tan(A-B)

if you dont understand try solving backwords from above formula of

tan(A+B)/tan(A-B)

Happy to help my friend anyway do you have exam tomorrow :)

Answer 2

Answer:

Step-by-step explanation:

sin(a+b) cos(a-b) /cos(a+b) sin(a-b)

=x+1/x-1

By componando divindo

sin(a+b) cos(a-b) +cos(a+b) sin(a-b) /

sin(a+b) cos(a-b) - cos(a+b) sin(a-b) =x+1-1+x/x-x+1+1

Sin(a+b+a-b)/sin(a+b-a+b)=2x/2

Sin2a/sin2b=x/1

Sin2a=2sin2b

By reviseing it we can proof it