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Question: If\(\frac{\tan\alpha - i\left( \sin\frac{\alpha}{2} + \cos\frac{\alpha}{2} \right)}{1 + 2i\sin\frac{...

Iftan⁡α−i(sin⁡α2+cos⁡α2)1+2isin⁡α2\frac{\tan\alpha - i\left( \sin\frac{\alpha}{2} + \cos\frac{\alpha}{2} \right)}{1 + 2i\sin\frac{\alpha}{2}} is purely imaginary, then α is given by

A

nπ + π2\frac{\pi}{2}

B

nπ – π4\frac{\pi}{4}

C

2nπ

D

None of these

Answer

2nπ

Explanation

Solution

Sol.

Re( tan⁡α−i(sin⁡α2+cos⁡α2)1+2isin⁡α2)\text{Re}\left( \ \frac{\tan\alpha - i\left( \sin\frac{\alpha}{2} + \cos\frac{\alpha}{2} \right)}{1 + 2i\sin\frac{\alpha}{2}} \right) = 0

or Re( tan⁡α−i(sin⁡α2+cos⁡α2)(1−2isin⁡α2)1+4sin⁡2α2)\text{Re}\left( \ \frac{\tan\alpha - i\left( \sin\frac{\alpha}{2} + \cos\frac{\alpha}{2} \right)\left( 1 - 2i\sin\frac{\alpha}{2} \right)}{1 + 4\sin^{2}\frac{\alpha}{2}} \right)

= 0 ⇒ tanα = 2 sin α2(sin⁡α2+cos⁡α2)\frac{\alpha}{2}\left( \sin\frac{\alpha}{2} + \cos\frac{\alpha}{2} \right)

⇒ tanα = 2 sin2 α2\frac{\alpha}{2} + sinα ⇒ sin⁡αcos⁡α\frac{\sin\alpha}{\cos\alpha}= sinα + 1 – cosα

⇒ sinα = sinα.cosα + cosα – cos2α

⇒ sinα (1– cosα) = cosα (1 – cosα)

⇒ sinα = cosα, cosα = 1

⇒ α = nπ +π4\frac{\pi}{4}, α = 2nπ (n∈I)