Question

If $0<x<\frac{\pi }{2}$, then the minimum value of $\frac{{\cos }^{4}x}{{\sin }^{2}x}+\frac{{\sin }^{4}x}{{\cos }^{2}x}$ is

• A. $\sqrt{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $1$

Answer 1

Answer: (D)

$1$

we know that, $A.M.\ge G.M.$

$⟹\frac{a+b}{2}\ge \sqrt{ab}$

let $a=\frac{co{s}^{4}x}{si{n}^{2}x}$ and $b=\frac{si{n}^{4}x}{co{s}^{2}x}$

therefore, $\frac{\frac{co{s}^{4}x}{si{n}^{2}x}+\frac{si{n}^{4}x}{co{s}^{2}x}}{2}\ge \sqrt{\frac{co{s}^{4}x}{si{n}^{2}x}\ast \frac{si{n}^{4}x}{co{s}^{2}x}}$

$⟹\frac{co{s}^{4}x}{si{n}^{2}x}+\frac{si{n}^{4}x}{co{s}^{2}x}\ge 2\ast \sqrt{si{n}^{2}x\ast co{s}^{2}x}$

$⟹\frac{co{s}^{4}x}{si{n}^{2}x}+\frac{si{n}^{4}x}{co{s}^{2}x}\ge 2\ast sinx\ast cosx$

$⟹\frac{co{s}^{4}x}{si{n}^{2}x}+\frac{si{n}^{4}x}{co{s}^{2}x}\ge sin2x$

for $0<x<\frac{\pi }{2}$ we get $0<sinx<1$

multiplying 2 we get $0<2x<\pi ⟹0<sinx\le 1$

therefore by substituting the maximum value of $sin2x$ we get

$\frac{co{s}^{4}x}{si{n}^{2}x}+\frac{si{n}^{4}x}{co{s}^{2}x}\ge 1$

therefore the minimum value of $\frac{co{s}^{4}x}{si{n}^{2}x}+\frac{si{n}^{4}x}{co{s}^{2}x}$ is $1$