The value of {tan ^2}theta + {cot ^2}theta is | vedclass

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(A) We know that for any real number $x 
eq 0$,the square of a real number is always non-negative,so $(x - \frac{1}{x})^2 \ge 0$.Expanding this,we ge

Vedclass · Accepted Answer

$\ge 2$

Vedclass · Answer

(A) We know that for any real number $x 
eq 0$,the square of a real number is always non-negative,so $(x - \frac{1}{x})^2 \ge 0$.Expanding this,we get $x^2 + \frac{1}{x^2} - 2 \ge 0$,which implies $x^2 + \frac{1}{x^2} \ge 2$.Let $x = 	an 	heta$. Since $\cot 	heta = \frac{1}{	an 	heta}$,substituting this into the inequality gives $	an^2 	heta + \cot^2 	heta \ge 2$.Thus,the expression is always greater than or equal to $2$.

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

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