Question

Prove \[{{\sec }^{2}}x+\cos e{{c}^{2}}x\ge 4\].

Answer 1

Hint: In the following question, we have to prove that \[{{\sec }^{2}}x+\cos e{{c}^{2}}x\ge 4\]. We know that trigonometric functions have certain identities which can be used here: \[{{\sec }^{2}}x=1+{{\tan }^{2}}x\]
\[\cos e{{c}^{2}}x=1+{{\cot }^{2}}x\]
When we use these identities along with the inequality with the concepts of Arithmetic mean and geometric mean, then we can prove the above statement.

Complete step by step solution:
We have the expression as follows: \[{{\sec }^{2}}x+\cos e{{c}^{2}}x\]
 This can be further written as follows: \[1+{{\tan }^{2}}x+1+{{\cot }^{2}}x\]
Here we use \[{{\sec }^{2}}x=1+{{\tan }^{2}}x\] and \[\cos e{{c}^{2}}x=1+{{\cot }^{2}}x\]
We get, \[{{\sec }^{2}}x+\cos e{{c}^{2}}x=\]\[2+{{\tan }^{2}}x+{{\cot }^{2}}x\]
Now we know that arithmetic mean is greater than geometric mean that is \[AM\ge GM\]
Therefore, \[\dfrac{{{\tan }^{2}}x+{{\cot }^{2}}x}{2}\ge \sqrt{{{\tan }^{2}}x{{\cot }^{2}}x}\]
$\Rightarrow$\[\dfrac{{{\tan }^{2}}x+{{\cot }^{2}}x}{2}\ge \sqrt{\dfrac{1}{{{\cot }^{2}}x}\times {{\cot }^{2}}x}\]
$\Rightarrow$\[\dfrac{{{\tan }^{2}}x+{{\cot }^{2}}x}{2}\ge 1\]
$\Rightarrow$\[{{\tan }^{2}}x+{{\cot }^{2}}x\ge 2\]
Now on adding two on both sides of the inequality:
$\Rightarrow$\[{{\tan }^{2}}x+{{\cot }^{2}}x+2\ge 2+2\]
$\Rightarrow$\[{{\tan }^{2}}x+{{\cot }^{2}}x+2\ge 4\]
$\Rightarrow$\[{{\tan }^{2}}x+1+{{\cot }^{2}}x+1\ge 4\]
So, \[{{\sec }^{2}}x+\cos e{{c}^{2}}x\ge 4\]
Hence prove that \[{{\sec }^{2}}x+\cos e{{c}^{2}}x\ge 4\].

Additional Information: We are given the expression: \[{{\sec }^{2}}x+\cos e{{c}^{2}}x\]
We know that in trigonometry \[\sec x=\dfrac{1}{\cos x}\]and \[\cos ecx=\dfrac{1}{\sin x}\]
So, \[{{\sec }^{2}}x=\dfrac{1}{{{\cos }^{2}}x}\]and \[\cos e{{c}^{2}}x=\dfrac{1}{si{{n}^{2}}x}\]
Therefore, the expression can be rewritten as: \[\dfrac{1}{si{{n}^{2}}x}+\dfrac{1}{{{\cos }^{2}}x}\]
\[\Rightarrow \dfrac{si{{n}^{2}}x+{{\cos }^{2}}x}{si{{n}^{2}}x{{\cos }^{2}}x}\]
\[\Rightarrow \dfrac{1}{{{\cos }^{2}}xsi{{n}^{2}}x}\] Here, we use identity \[{{\cos }^{2}}x+si{{n}^{2}}x=1\]
We know that\[AM\ge GM\], that is Arithmetic mean is greater than Geometric Mean
So, \[\dfrac{1}{2{{\cos }^{2}}xsi{{n}^{2}}x}\ge \sqrt{\dfrac{1}{{{\cos }^{2}}xsi{{n}^{2}}x}}\]
\[\Rightarrow {{\left( \dfrac{1}{2{{\cos }^{2}}xsi{{n}^{2}}x} \right)}^{2}}\ge \dfrac{1}{{{\cos }^{2}}xsi{{n}^{2}}x}\]
\[\Rightarrow \dfrac{1}{4{{\cos }^{4}}xsi{{n}^{4}}x}\ge \dfrac{1}{{{\cos }^{2}}xsi{{n}^{2}}x}\]
\[\dfrac{{{\cos }^{2}}xsi{{n}^{2}}x}{{{\cos }^{4}}xsi{{n}^{4}}x}\ge 4\]
\[\dfrac{1}{{{\cos }^{2}}xsi{{n}^{2}}x}\ge 4\]
\[\dfrac{{{\cos }^{2}}x+si{{n}^{2}}x}{{{\cos }^{2}}xsi{{n}^{2}}x}\ge 4\]
\[\dfrac{1}{{{\cos }^{2}}x}+\dfrac{1}{{{\sin }^{2}}x}\ge 4\]
Hence, \[{{\sec }^{2}}x+\cos e{{c}^{2}}x\ge 4\]

Note: Arithmetic Mean: When three numbers say a, b and c are in Arithmetic Progression such that \[b-a=c-b=d\]= common difference, then
Arithmetic mean is equal to: \[\dfrac{a+c}{2}=b\]
Geometric Mean: if three terms are in geometric progression say a, b, c such that \[\dfrac{b}{a}=\dfrac{c}{b}=r\] r=common ratio then middle term is called geometric mean of first and last number. Therefore geometric mean is equal to \[b=\sqrt{ac}\]
Always relate one chapter from the other chapter like in this question. We use the concept of geometric mean and arithmetic mean in the trigonometric proof. Basic formulas of trigonometry you should remember. This type of result is generally asked in JEE mains examination. In Inequality equations if we multiply with a negative number on both sides then the sign of inequality will change.