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If α, β are the solutions of a tanθ+bsecθ=c, then tan (α+β)=

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a
2aca2c2
b
2acc2a2
c
2aca2+c2
d
aca2+c2

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detailed solution

Correct option is A

We have

atanθ+bsecθ=cbsecθ=catanθb2sec2θ=c2+a2tan2θ2actanθb21+tan2θ=c2+a2tan2θ2actanθtan2θb2a2+2actanθ+b2c2=0

Since tan α and tan β are roots of this equation

 tanα+tanβ=2acb2a2 and tanαtanβ=b2c2b2a2

Now,

tan(α+β)=tanα+tanβ1tanαtanβ=2acb2a21b2c2b2a2=2aca2c2

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