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Question: 4] b) If tan($\theta$+i$\omega$) = tan$\alpha$ + isec$\alpha$ Prove that
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$1) e^{2φ} = \frac{cotα}{22} 2θ = nπ + \frac{π}{2} +α.$

Subject:- Applied Mathematics

Marks:- 3

Mumbai Unversity>FE>Sem1>Applied Maths1

m1(81) • 8 views
 modified 4 days ago  • written 4 days ago by
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$tan(θ+iφ) = tanα + isecα∴ tan(θ-iφ) = tanα - isecα$ $∴ tan2θ = tan[(θ+iφiφ+ (θ-iφ)] = \frac{(tan⁡(θ+iφ) +tan⁡(θ-iφ))}{(1-tan⁡(θ+iφ) tan⁡(θ-iφ) )}$ $=\frac{(tan⁡(θ+isecα) +tan⁡(θ-isecα))}{(1-tan⁡(θ+isecα) tan⁡(θ-isecα) )}$=$\frac{ 2tanα}{(1-(tan)^2 α+(sec)^2 α )}$=$\frac{2tanα}{(-2(tan)^2 α)-cotα}$ =$tan(\frac{π}{2}+ α)$

$∴ 2θ= nπ + \frac{π}{2}+ α. (general value)$.

$Again tan(2iφ) = tan[(θ+iφ) - (θ-iφ)]$

$= \frac{(tan⁡(θ+isecα)-tan⁡(θ-isecα))}{(1+tan⁡(θ+isecα) tan⁡(θ-isecα) )}$

$∴ itanh2φ = \frac{2isecα}{(2(sec)^2 α)}= icosα∴ tanh2φ = cosα$

$∴ 2φ = tanh^{-1} (cosα) (cosα\frac{1}{2}log[\frac{(1+cosα)}{(1-cosα)}] = 1/2log[\frac{(2(cos)^2 (α/2))}{(2(sin)^2 (α/2) )}]ogcotα/2$