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Question

Question: If \(\alpha + \beta + \gamma = 2\pi,\) then...

If α+β+γ=2π,\alpha + \beta + \gamma = 2\pi, then

A

tan⁡α2+tan⁡β2+tan⁡γ2=tan⁡α2tan⁡β2tan⁡γ2\tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \tan\frac{\gamma}{2} = \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\tan\frac{\gamma}{2}

B

tan⁡α2tan⁡β2+tan⁡β2tan⁡γ2+tan⁡γ2tan⁡α2=1\tan\frac{\alpha}{2}\tan\frac{\beta}{2} + \tan\frac{\beta}{2}\tan\frac{\gamma}{2} + \tan\frac{\gamma}{2}\tan\frac{\alpha}{2} = 1

C

tan⁡α2+tan⁡β2+tan⁡γ2=−tan⁡α2tan⁡β2tan⁡γ2\tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \tan\frac{\gamma}{2} = - \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\tan\frac{\gamma}{2}

D

None of these

Answer

tan⁡α2+tan⁡β2+tan⁡γ2=tan⁡α2tan⁡β2tan⁡γ2\tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \tan\frac{\gamma}{2} = \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\tan\frac{\gamma}{2}

Explanation

Solution

We have α+β+γ=2π⇒α2+β2+γ2=π\alpha + \beta + \gamma = 2\pi \Rightarrow \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} = \pi

⇒tan⁡(α2+β2+γ2)=tan⁡π=0\Rightarrow \tan\left( \frac{\alpha}{2} + \frac{\beta}{2} + \frac{\gamma}{2} \right) = \tan\pi = 0

⇒tan⁡α2+tan⁡β2+tan⁡γ2−tan⁡α2tan⁡β2tan⁡γ2=0\Rightarrow \tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \tan\frac{\gamma}{2} - \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\tan\frac{\gamma}{2} = 0

⇒tan⁡α2+tan⁡β2+tan⁡γ2=tan⁡α2tan⁡β2tan⁡γ2\Rightarrow \tan\frac{\alpha}{2} + \tan\frac{\beta}{2} + \tan\frac{\gamma}{2} = \tan\frac{\alpha}{2}\tan\frac{\beta}{2}\tan\frac{\gamma}{2}.