If theta is an acute angle and tan^{2} theta | vedclass

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(D) Given the equation: $	an^{2} 	heta + \frac{1}{	an^{2} 	heta} = 2$.Let $x = 	an^{2} 	heta$. Then the equation becomes $x + \frac{1}{x} = 2$.M

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(D) Given the equation: $	an^{2} 	heta + \frac{1}{	an^{2} 	heta} = 2$.Let $x = 	an^{2} 	heta$. Then the equation becomes $x + \frac{1}{x} = 2$.Multiplying by $x$,we get $x^{2} + 1 = 2x$,which simplifies to $x^{2} - 2x + 1 = 0$.This is a perfect square: $(x - 1)^{2} = 0$,so $x = 1$.Substituting back,$	an^{2} 	heta = 1$.Since $	heta$ is an acute angle,$	an 	heta = 1$.Therefore,$	heta = 45^{\circ}$.

If $\theta$ is an acute angle and $\tan^{2} \theta + \frac{1}{\tan^{2} \theta} = 2$,then the value of $\theta$ is (in $^{\circ}$)

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