The correct options are A e^2ϕ=α2 B 2θ=nπ+π2+α We have tan(θ+iϕ)=tanα+iα and tan(θ iϕ)=tanα iα Now, tan(θ+iϕ+θ iϕ)=tan2θ =tan(θ+i ...

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Sign of Trigonometric Ratios in Different Quadrants

If tanθ+iϕ=...

Question

If $tan (θ + i ϕ) = tan α + i sec α$ then

e^{2 ϕ} = cot \frac{α}{2}

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2 θ = n π + \frac{π}{2} + α

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e^{2 ϕ} = - cot \frac{α}{2}

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2 θ = n π - \frac{π}{2} + α

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Solution

The correct options are
A $e^{2 ϕ} = cot \frac{α}{2}$
B $2 θ = n π + \frac{π}{2} + α$
We have $tan (θ + i ϕ) = tan α + i sec α$
and $tan (θ - i ϕ) = tan α - i sec α$
Now, $t a n (θ + i ϕ + θ - i ϕ) = tan 2 θ$
$= \frac{tan (θ + i ϕ) + tan (θ - i ϕ)}{1 - (tan α + i sec α) (tan α - i sec α)}$
$\frac{tan α + i sec α + tan α - i sec α}{1 - (tan α + i sec α) (tan α - i sec α)}$
$= \frac{2 tan α}{1 - ({tan}^{2} α - i^{2} {s e c}^{2} α)}$
As $i^{2} = - 1$
$\Rightarrow \frac{2 tan α}{1 - {tan}^{2} α - {sec}^{2} α}$
We know that $1 - {sec}^{2} α = - {tan}^{2} α$
$\Rightarrow \frac{2 tan α}{- 2 {tan}^{2} α}$
$- cot α$
We have $tan 2 θ = - cot α$
$\Rightarrow tan 2 θ = tan (\frac{π}{2} + α)$
$\Rightarrow 2 θ = n π + \frac{π}{2} + α$
Now, $tan 2 i ϕ = tan [(θ + i ϕ) - (θ - i ϕ)]$
$= \frac{tan (θ + i ϕ) + tan (θ - i ϕ)}{1 + tan (θ + i ϕ) tan (θ - i ϕ)}$
$= \frac{tan α + i sec α - tan α + i sec α}{1 + (tan α + i sec α) (tan α - i sec α)}$
$= \frac{2 i sec α}{1 + {tan}^{2} α - i^{2} {sec}^{2} α}$
$= \frac{2 i sec α}{2 {sec}^{2} α}$
$= \frac{i}{sec α}$
$= i cos α$
$\Rightarrow tan h 2 ϕ = cos α$
$2 ϕ = tan h^{- 1} (cos α)$
$2 ϕ = \frac{1}{2} log (\frac{1 + cos α}{1 - cos α})$
$2 ϕ = \frac{1}{2} log ⎛ ⎝ \frac{2 {cos}^{2} \frac{α}{2}}{2 {sin}^{2} \frac{α}{2}} ⎞ ⎠$
$\Rightarrow 2 ϕ = \frac{1}{2} log {cot}^{2} \frac{α}{2}$
or $2 ϕ = \frac{2}{2} log cot \frac{α}{2}$
or $2 ϕ = log cot \frac{α}{2}$
or $e^{2 ϕ} = cot \frac{α}{2}$

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