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JEE MATHEMATICS MCQ: COMPOUND ANGLES

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If   α+β+γ=2π\;\alpha + \beta + \gamma = 2\pi, then
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Correct option: A
tan⁡  α2+tan⁡  β2+tan⁡  γ2=tan⁡  α2  tan⁡  β2  tan⁡  γ2\tan \,\,\frac{\mathrm{\alpha }}{2} + \tan \,\,\frac{\mathrm{\beta }}{2} + \tan \,\,\frac{\mathrm{\gamma }}{2} = \tan \,\,\frac{\mathrm{\alpha }}{2}\,\,\tan \,\,\frac{\mathrm{\beta }}{2}\,\,\tan \,\,\frac{\mathrm{\gamma }}{2}
Option A is correct because when α+β+γ=2π\alpha + \beta + \gamma = 2\pi, the sum of the tangents of half angles equals their product.

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 If A,B,CA,B,C are angles of a triangle, then the minimum value of tan⁡2  A2+tan⁡2  B2+tan⁡2  C2{\tan ^2}\,\,\frac{A}{2} + {\tan ^2}\,\,\frac{B}{2} + {\tan ^2}\,\,\frac{C}{2}, is If A>0,   B>0A > 0,\,\;B > 0 and A+B=π3A + B = \frac{\pi }{3}, then the maximum value of tan⁡A  tan⁡B\tan A\,\,\tan B, is
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