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Question

If tan(A+B) =3 and sin(A-B) = 12 Where A and B are acute angles.

Then, Cos2A + Sin2B =__

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Solution

Tan (A+B)=3

And we know that tan 60 =3

So, A+B =60………………………………………………….(eq. 1)

and, sin(A-B)=12

We know that sin 30=12

So, A-B=30……………………………………………………(eq. 2)

Adding eq.1 and eq.2, we get

2A=90

A=45 & B= 15


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