Question

Prove that:

3 (\cot^{2} θ + \tan^{2} θ) - 8 (\cot θ + \tan θ) + 10 \geq 0

for real

θ \neq n π, n π + \frac{π}{2}

where

n \in Z

.

Accepted Answer

Let $\cot θ + \tan θ = p$

Then, $\cot^{2} θ + \tan^{2} θ = (\cot θ + \tan θ)^{2} - 2 \cot θ \cdot \tan θ$

$= p^{2} - 2$

$∴$ The given equation can be written as

$3 (p^{2} - 2) - 8 p + 10$

$= 3 p^{2} - 8 p + 4$

$= 3 (p^{2} - \frac{8}{3} p) + 4$
$= 3 (p^{2} - 2 \cdot \frac{4}{3} p + \frac{16}{9}) + 4 - 3 \cdot \frac{16}{9}$
$\Rightarrow 3 (\cot^{2} θ + \tan^{2} θ) - 8 (\cot θ + \tan θ) + 10 = 3 {(p - \frac{4}{3})}^{2} - \frac{4}{3} . . . (1)$

If $p < 0,$ clearly $3 {(p - \frac{4}{3})}^{2} - \frac{4}{3} > 3 {(\frac{4}{3})}^{2} - \frac{4}{3}$ that is the expression is positive.

Also, we know that, if $θ$ is an angle in the second or the fourth quadrant then $\tan θ$ and $c o t$ $θ$ are both negative and so is $\cot θ + \tan θ$ .

Thus, $p$ is negative. Thus, it can be said that the given expression is positive when $θ$ is an angle in the second or the fourth quadrant.

If $θ$ is in first or third quadrant, both $\cot θ$ and $\tan θ$ will be positive; So, $A . M . \geq G . M .$

$\Rightarrow \frac{\cot θ + \tan θ}{2} \geq \sqrt{\cot θ \cdot \tan θ} i . e ., p \geq 2$

When $p \geq 2,$ from $(1)$

$3 (\cot^{2} θ + \tan^{2} θ) - 8 (\cot θ + \tan θ) + 10 \geq 3 {(2 - \frac{4}{3})}^{2} - \frac{4}{3} = 3 \cdot \frac{4}{9} - \frac{4}{3} = 0$

Thus, for all $θ$ except quadrant angles , the given expression is greater than or equal to $0$ .

Amit M Agarwal Solutions for Chapter: Trigonometric Equations and Inequations, Exercise 1: Exercise on Level

Amit M Agarwal Mathematics Solutions for Exercise - Amit M Agarwal Solutions for Chapter: Trigonometric Equations and Inequations, Exercise 1: Exercise on Level

Questions from Amit M Agarwal Solutions for Chapter: Trigonometric Equations and Inequations, Exercise 1: Exercise on Level with Hints & Solutions

Learn and Prepare for any exam you want !