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Question

General value of θ satisfying the equation tan2θ+sec2θ=1 is

A
mπ,nπ+π3
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B
mπ,nπ±π3
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C
mπ,nπ±π6
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D
None of these
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Solution

The correct option is D mπ,nπ±π3
By using, sec2θ=1cos2θ=1+tan2θ1tan2θ
We can write the given equation as
tan2θ+1+tan2θ1tan2θ
tan2θ+1+tan2θ1tan2θ=1
tan2θ(1tan2θ)+1+tan2θ=1tan2θ
3tan2θtan4θ=0
tan2θ(3tan2θ)=0
tanθ=0ortanθ=±3
Now, tanθ=0θ=mπ, where m is an integer and tanθ=±3tan[±π3]
θ=nπ±π3 where n is an integer
Thus, θ=mπ,nπ±π3

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