Question

If $p={{\tan }^{2}}x+{{\cot }^{2}}x$, then which of the following is correct?
[a] $p\le 2$
[b] $p\ge 2$
[c] $p<2$
[d] $p>2$

Answer 1

Hint: Put $\tan x = t$ and use the fact that $\cot x=\dfrac{1}{\tan x}$. Hence prove that the given expression is equal to ${{t}^{2}}+\dfrac{1}{{{t}^{2}}}$. Subtract and add 2 to the expression and use the algebraic identity ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$. Hence prove that the given expression is equal to ${{\left( t-\dfrac{1}{t} \right)}^{2}}+2$. Use the fact that $\forall x\in \mathbb{R},{{x}^{2}}\ge 0$. Hence prove that ${{\left( t-\dfrac{1}{t} \right)}^{2}}+2\ge 2$. Hence find the range of p. Alternatively, use AM-GM inequality, i.e. if $a\ge 0,b\ge 0$, then $\dfrac{a+b}{2}\ge {{\left( ab \right)}^{\dfrac{1}{2}}}$. Hence find the range of p. Alternatively, differentiate both sides and hence find the range of p.

Complete step-by-step answer:
Put tanx = t.
Now, we have
$p={{\tan }^{2}}x+{{\cot }^{2}}x$
We know that $\cot x=\dfrac{1}{\tan x}$
Using the above identity, we get
$p={{\tan }^{2}}x+\dfrac{1}{{{\tan }^{2}}x}$
Hence, we have
$p={{t}^{2}}+\dfrac{1}{{{t}^{2}}}$
Subtracting and adding 2 on RHS, we get
$p={{t}^{2}}+\dfrac{1}{{{t}^{2}}}-2+2$
Now, we know that $2\left( t \right)\left( \dfrac{1}{t} \right)=2$
Hence, we have
$p={{t}^{2}}+\dfrac{1}{{{t}^{2}}}-2\left( t \right)\left( \dfrac{1}{t} \right)+2$
We know that ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$
Using the above algebraic identity, we get
$p={{\left( t-\dfrac{1}{t} \right)}^{2}}+2$
Now, we know that $\forall x\in \mathbb{R},{{x}^{2}}\ge 0$
Hence, we have
$\forall t\in \mathbb{R},{{\left( t-\dfrac{1}{t} \right)}^{2}}\ge 0$
Adding 2 on both sides, we get
${{\left( t-\dfrac{1}{t} \right)}^{2}}+2\ge 2$
Hence, we have
$p\ge 2$
Hence option [b] is correct.

Note: Alternative Solution:
We know that if $a\ge 0,b\ge 0$, then the arithmetic mean of a and b is greater or equal to the geometric mean of a and b, i.e. $\dfrac{a+b}{2}\ge {{\left( ab \right)}^{\dfrac{1}{2}}}$
Put $a={{\tan }^{2}}x,b={{\cot }^{2}}x$, we get
$\dfrac{{{\tan }^{2}}x+{{\cot }^{2}}x}{2}\ge {{\left( {{\tan }^{2}}x{{\cot }^{2}}x \right)}^{\dfrac{1}{2}}}$
We know that $\cot x=\dfrac{1}{\tan x}\Rightarrow \cot x\tan x=1$
Hence, we have
$\dfrac{{{\tan }^{2}}x+{{\cot }^{2}}x}{2}\ge 1$
Multiplying both sides by 2, we get
${{\tan }^{2}}x+{{\cot }^{2}}x\ge 2$
Hence, we have
$p\ge 2$
Hence option [b] is correct.