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Question

If sin(α+β)=1 and sin(αβ)=12, where 0α,βπ2, then find the values of tan(α+2β) and tan(2α+β).

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Solution

We have
sin(α+β)=1
sin(αβ)=sinπ2
α+β=π2 .....(i)
and, sinαβ=12
sin(αβ)=sinπ6
αβ=π6 ....(ii)
Adding equations (i) and (ii), we get
2α=π2+π6=4π6=2π3
α=π3
Putting α=π3 in equation (i), we get
π3+β=π2
β=π2π3
β=3π2π6=π6
β=π6
Now, tan(α+2β)=tan(π3+2×π6)
=tan(π3+π3)
=tan2π3
=tan(π2+π6)
=cotπ6
=3
tan(α+23)=3
and, tan(2α+β)=tan(2×π3+π6)
=tan(2π3+π6)
=tan(4π+π6)
=tan(5π6)
=(π2+π3)
=cotπ3
=13
tan(2α+β)=13

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