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Question: If \(A + B + C = \pi,\) then \(\tan ^ { 2 } \frac { A } { 2 } + \tan ^ { 2 } \frac { B } { 2 } +\) \...

If A+B+C=π,A + B + C = \pi, then tan⁡2A2+tan⁡2B2+\tan ^ { 2 } \frac { A } { 2 } + \tan ^ { 2 } \frac { B } { 2 } + tan⁡2C2\tan^{2}\frac{C}{2} is always

A

≤1\leq 1

B

≥1\geq 1

C

= 0

D

= 1

Answer

≥1\geq 1

Explanation

Solution

tan⁡(A2+B2+C2)=S1−S31−S2=tan⁡π2=∞\tan\left( \frac{A}{2} + \frac{B}{2} + \frac{C}{2} \right) = \frac{S_{1} - S_{3}}{1 - S_{2}} = \tan\frac{\pi}{2} = \infty

∴S2=1\therefore S_{2} = 1 or xy+yz+zx=1xy + yz + zx = 1, where x=tan⁡A2x = \tan\frac{A}{2}etc.

Now (x−y)2+(y−z)2+(z−x)2≥0(x - y)^{2} + (y - z)^{2} + (z - x)^{2} \geq 0or

2∑x2−2∑xy≥0⇒∑x2≥12\sum x^{2} - 2\sum xy \geq 0 \Rightarrow \sum x^{2} \geq 1. {∵∑xy=1}\{\because\sum xy = 1\}