Question

If A + B + C = $\pi$ then $ta{n}^2\frac{A}{2}$ + $ta{n}^2\frac{B}{2}$ + $ta{n}^2\frac{C}{2}$ is always

• A. $\le$ 1
• B. $\ge$ 1
• C. = 0
• D. = 1

Answer 1

The correct option is B

$\ge$ 1

Given A + B + C = $\pi$

All the angles given with $\frac{A}{2}$, $\frac{B}{2}$ and $\frac{C}{2}$ form

So, $\frac{A+B+C}{2}$ = $\frac{\pi }{2}$

$\frac{A}{2}$ + $\frac{B}{C}$ = $\frac{\pi }{2}$ - $\frac{C}{2}$

tan $(\frac{A}{2}+\frac{B}{C})$ + tan $(\frac{\pi }{2}-\frac{C}{2})$

$\frac{tan\frac{A}{2}+tan\frac{B}{2}}{1-tan\frac{A}{2}.tan\frac{B}{2}}$ = $cot\frac{C}{2}$ = $\frac{1}{tan\frac{C}{2}}$

$tan\frac{A}{2}$ . $tan\frac{C}{2}$ + $tan\frac{B}{2}$ . $tan\frac{C}{2}$ = 1 - $tan\frac{A}{2}$ . $tan\frac{B}{2}$

$tan\frac{A}{2}$ . $tan\frac{C}{2}$ + $tan\frac{A}{2}$ . $tan\frac{B}{2}$ + $tan\frac{A}{2}$ . $tan\frac{B}{2}$ = 1

Let x = $tan\frac{A}{2}$

y = $tan\frac{B}{2}$

z = $tan\frac{C}{2}$

then

xy + yz + zx = 1

We need to find the value of $ta{n}^2\frac{A}{2}+ta{n}^2\frac{B}{2}+ta{n}^2\frac{C}{2}or{x}^2+y^2+z^2$

We know

$ta{n}^2\frac{A}{2}+ta{n}^2\frac{B}{2}+ta{n}^2\frac{C}{2}or{x}^2+y^2+z^2$

$x^2$ + $y^2$ - 2xy + $y^2$ + $z^2$ - 2yz + $z^2$ + $x^2$ - 2xz ≥ 0

2($x^2$ + $y^2$ + $z^2$) - 2(xy + yz + zx) $\ge$ 0

($x^2$ + $y^2$ + $z^2$) - (xy + yz + zx) $\ge$ 0

Given xy + yz + zx = 1

$x^2$ + $y^2$ + $z^2$ - 1 $\ge$ 0

$x^2$ + $y^2$ + $z^2$ $\ge$ 1

Replacing x = $tan\frac{A}{2}$

Replacing x = $tan\frac{A}{2}$

y = $tan\frac{B}{2}$

z = $tan\frac{C}{2}$

So $ta{n}^2\frac{C}{2}$ + $ta{n}^2\frac{C}{2}$ + $ta{n}^2\frac{C}{2}$ $\ge$ 1