The minimum value of 27 tan ^2 theta + 3 cot | vedclass

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(B) We know that for any positive real numbers $a$ and $b$,the Arithmetic Mean is greater than or equal to the Geometric Mean,i.e.,$\frac{a+b}{2} \geq

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(B) We know that for any positive real numbers $a$ and $b$,the Arithmetic Mean is greater than or equal to the Geometric Mean,i.e.,$\frac{a+b}{2} \geq \sqrt{ab}$.Let $a = 27 	an ^2 	heta$ and $b = 3 \cot ^2 	heta$.Then,$\frac{27 	an ^2 	heta + 3 \cot ^2 	heta}{2} \geq \sqrt{27 	an ^2 	heta \cdot 3 \cot ^2 	heta}$.$\frac{27 	an ^2 	heta + 3 \cot ^2 	heta}{2} \geq \sqrt{81 	an ^2 	heta \cdot \cot ^2 	heta}$.Since $	an 	heta \cdot \cot 	heta = 1$,we have $\frac{27 	an ^2 	heta + 3 \cot ^2 	heta}{2} \geq \sqrt{81 \cdot 1}$.$\frac{27 	an ^2 	heta + 3 \cot ^2 	heta}{2} \geq 9$.$27 	an ^2 	heta + 3 \cot ^2 	heta \geq 18$.Thus,the minimum value is $18$.

The minimum value of $27 \tan ^2 \theta + 3 \cot ^2 \theta$ is

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