Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

Since $\frac{A}{2}+\frac{B}{2}+\frac{C}{2}=\frac{\pi }{2},\tan \left(\frac{A}{2}+\frac{B}{2}\right)=\tan \left(\frac{\pi }{2}-\frac{C}{2}\right)$ gives $\tan \frac{A}{2}\tan \frac{B}{2}+\tan \frac{B}{2}\tan \frac{C}{2}+\tan \frac{C}{2}\tan \frac{A}{2}=1$. Then ${\tan }^2\frac{A}{2}+{\tan }^2\frac{B}{2}+{\tan }^2\frac{C}{2}-1=\frac{1}{2}\left[{\left(\tan \frac{A}{2}-\tan \frac{B}{2}\right)}^2+{\left(\tan \frac{B}{2}-\tan \frac{C}{2}\right)}^2+{\left(\tan \frac{C}{2}-\tan \frac{A}{2}\right)}^2\right]\ge 0$. Hence the minimum value is 1.