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Question

If tan(θ+iϕ)=tanα+isecα then

A
e2ϕ=cotα2
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B
2θ=nπ+π2+α
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C
e2ϕ=cotα2
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D
2θ=nππ2+α
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Solution

The correct options are
A e2ϕ=cotα2
B 2θ=nπ+π2+α
We have tan(θ+iϕ)=tanα+isecα
and tan(θiϕ)=tanαisecα
Now, tan(θ+iϕ+θiϕ)=tan2θ
=tan(θ+iϕ)+tan(θiϕ)1(tanα+isecα)(tanαisecα)
tanα+isecα+tanαisecα1(tanα+isecα)(tanαisecα)
=2tanα1(tan2αi2sec2α)
As i2=1
2tanα1tan2αsec2α
We know that 1sec2α=tan2α
2tanα2tan2α
cotα
We have tan2θ=cotα
tan2θ=tan(π2+α)
2θ=nπ+π2+α
Now, tan2iϕ=tan[(θ+iϕ)(θiϕ)]
=tan(θ+iϕ)+tan(θiϕ)1+tan(θ+iϕ)tan(θiϕ)
=tanα+isecαtanα+isecα1+(tanα+isecα)(tanαisecα)
=2isecα1+tan2αi2sec2α
=2isecα2sec2α
=isecα
=icosα
tanh2ϕ=cosα
2ϕ=tanh1(cosα)
2ϕ=12log(1+cosα1cosα)
2ϕ=12log2cos2α22sin2α2
2ϕ=12logcot2α2
or 2ϕ=22logcotα2
or 2ϕ=logcotα2
or e2ϕ=cotα2

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