Question
Show that $\tan^2 \theta + \cot^2 \theta \geq 2$ for all $\theta \in R$.

Answer

$ \tan ^2 \theta+ \cot ^2 \theta=\tan ^2 \theta+\frac{1}{\tan ^2 \theta}$
$=(\tan \theta)^2+\left(\frac{1}{\tan \theta}\right)^2$
$=\left(\tan \theta-\frac{1}{\tan \theta}\right)^2+2 \tan \theta \cdot \frac{1}{\tan \theta}$
$\quad \ldots\left[\because a^2+b^2=(a-b)^2+2 a b\right]$
$=\left(\tan \theta-\frac{1}{\tan \theta}\right)^2+2 \geq 2 \text { for all } \theta \in R .$

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