Question

Sin (A+B)=1 and tan(A-B)=1/√3.FIND 3B+2A/3B-2A...​

Answer 1

Step-by-step explanation:

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Answer 2

Question:-

Sin (A+B)=1 and tan(A-B)=1/√3.FIND 3B+2A/3B-2A...

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Tan ( A + B ) = 1

So.. Tan ( A + B ) = Tan 45°

By Comparing , A + B = 45 ______equation 1

Now , Tan ( A - B ) = 1/√3

So.. Tan ( A - B ) = Tan 30°

By Comparing , A - B = 30 _______equation2

From equation 1 & equation2 ( On adding )

We get A = 37.5°

And B= 7.5°

OR

tan (a+b ) = 1 , hence a+b = 45° ...................(1)

tan (a - b) = 1/√3 , hence a-b = 30° .....................(2)

By adding eqn.(1) and (2), we get 2a = 75° or a = 37.5°

if we substitute a = 37.5° in eqn.(1), we get b = 7.5°

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