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Question: The real part of tan (a + ib) is –...

The real part of tan (a + ib) is –

A

2sin⁡2α2cos⁡2α+e2β+e−2β\frac{2\sin 2\alpha}{2\cos 2\alpha + e^{2\beta} + e^{- 2\beta}}

B

sin⁡2α2cos⁡α+e2β+e−2β\frac{\sin 2\alpha}{2\cos\alpha + e^{2\beta} + e^{- 2\beta}}

C

e2β+e−2βsin⁡2α\frac{e^{2\beta} + e^{- 2\beta}}{\sin 2\alpha}

D

None of these

Answer

2sin⁡2α2cos⁡2α+e2β+e−2β\frac{2\sin 2\alpha}{2\cos 2\alpha + e^{2\beta} + e^{- 2\beta}}

Explanation

Solution

Sol. tan (a + bi)

= sin⁡(α+βi)cos⁡(α+βi)\frac{\sin(\alpha + \beta i)}{\cos(\alpha + \beta i)} = 2sin⁡(α+βi).cos⁡(α−βi)2cos⁡(α+βi).cos⁡(α−βi)\frac{2\sin(\alpha + \beta i).\cos(\alpha - \beta i)}{2\cos(\alpha + \beta i).\cos(\alpha - \beta i)}

= sin⁡2α+sin⁡(2iβ)cos⁡2α+cos⁡(2iβ)\frac{\sin 2\alpha + \sin(2i\beta)}{\cos 2\alpha + \cos(2i\beta)}= sin⁡2α+ie2β−e−2β2cos⁡2α+e2β+e−2β2\frac{\sin 2\alpha + i\frac{e^{2\beta} - e^{- 2\beta}}{2}}{\cos 2\alpha + \frac{e^{2\beta} + e^{- 2\beta}}{2}}

=2sin⁡2α+i(e2β−e−2β)2cos⁡2α+e2β+e−2β\frac{2\sin 2\alpha + i(e^{2\beta} - e^{- 2\beta})}{2\cos 2\alpha + e^{2\beta} + e^{- 2\beta}}

\ Real part of tan (a + bi)

=2sin⁡2α2cos⁡2α+e2β+e−2β\frac{2\sin 2\alpha}{2\cos 2\alpha + e^{2\beta} + e^{- 2\beta}}.