Question

What is the least value of $\frac{25{\sec }^{4}x-50{\sec }^{2}x+74}{{\tan }^{2}x}$ ?

• A. $50$
• B. $70$
• C. $75$
• D. $90$

Answer 1

The correct option is B $70$

$\frac{25{\sec }^{4}x-50{\sec }^{2}x+74}{{\tan }^{2}x}=\frac{\left\{\right(5{\sec }^{2}x{)}^2-2\left(5\right)\left(5{\sec }^{2}x\right)+\left(5{)}^2\right\}+49}{{\tan }^{2}x}=\frac{25({\sec }^{2}x-1{)}^2}{{\tan }^{2}x}+\frac{49}{{\tan }^{2}x}=25{\tan }^{2}x+49{\cot }^{2}x=(5\tan x-7\cot x{)}^2+70\therefore min\{(5\tan x-7\cot x{)}^2+70\}=70)$