(A) We know that for any real number $x \neq 0$,$(x - \frac{1}{x})^2 \ge 0$.Expanding this,we get $x^2 + \frac{1}{x^2} - 2 \ge 0$,which implies $x^2 +

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(A) We know that for any real number $x \neq 0$,$(x - \frac{1}{x})^2 \ge 0$.Expanding this,we get $x^2 + \frac{1}{x^2} - 2 \ge 0$,which implies $x^2 + \frac{1}{x^2} \ge 2$.Let $x = \tan \theta$. Since $\cot \theta = \frac{1}{\tan \theta}$,we substitute $x = \tan \theta$ into the inequality:$\tan^2 \theta + \frac{1}{\tan^2 \theta} \ge 2$$\tan^2 \theta + \cot^2 \theta \ge 2$.Thus,the correct option is $A$.

Class 11 Mathematics Trigonometrical Ratios, Functions and Identities Maximum and minimum values of trigonometrical functions, Conditional trigonometrical identities

Easy

The value of $\tan^2 \theta + \cot^2 \theta$ is

A
$\ge 2$
B
$\le 2$
C
$\ge -2$
D
None of these

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Trigonometrical Ratios, Functions and Identities

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Maximum and minimum values of trigonometrical functions, Conditional trigonometrical identities

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