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Question

If α and β satisfies the equations 5sin2β=3sin2α and 3tanα=tanβ simultaneously, where 0<α<π, 0<β<π, α,βπ2, then the possible value(s) of tanα+tanβ is/are

A
4
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B
4
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C
2
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D
2
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Solution

The correct option is B 4
5sin2β=3sin2α
3tanα=tanβ(1)

5sin2β=3sin2α
52tanβ1+tan2β=32tanα1+tan2α
Using equation (1),
53tanα1+9tan2α=3tanα1+tan2α51+9tan2α=11+tan2α(tanα0)
5+5tan2α=1+9tan2α
tan2α=1
tanα=±1

When tanα=1
3tanα=tanβtanβ=3tanα+tanβ=1+3=4

When tanα=1
tanβ=3tanα+tanβ=13=4

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