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Question

If sin(AB)=110,cos(A+B)=229, fid the value of tan2A where A and B lies between 0 and π/4

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Solution

tan2A=tan[(A+B)+(AB)]=tan(A+B)+tan(AB)1tan(A+B)tan(AB) (i)
Given that, 0<A<π/4 and 0<B<π/4. Therefor,
0<A+B<π2
Also, π4<AB<π4 and sin(AB)=110=(+)ve
0<AB<π4
Now, sin(AB)=110
tan(AB)=13 (ii)
cos(A+B)=229
tan(AB)=52 (iii)
From Eqs. (i), (ii) and (iii), we get
tan2A=52+13152×13=176×61=17
Ans: 17

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