# Minimum value of f(θ)= a²sec²θ +b² cosec²θ using AM- GM inequality

If we take and function

$$f(θ)= \dfrac{a^2}{\cos^2\theta} + \dfrac{b^2}{\sin^2\theta}$$

And wish to find minimum value of function using AM -GM inequality

i.e. if we have two no. p & q Then $$AM \ge GM$$ $$\dfrac{p+q}{2} \ge \sqrt{pq}$$

Using here

$$\dfrac{a^2}{\cos^2\theta} + \dfrac{b^2}{\sin^2\theta} \ge 2 \sqrt { \dfrac{a²b²}{\sin^2 \theta \cos ^2 \theta}}$$

Multiply dividing by $$4$$ in square root

$$\dfrac{a^2}{\cos^2\theta} + \dfrac{b^2}{\sin^2\theta} \ge 2 \sqrt { \dfrac{4a²b²}{4\sin^2 \theta \cos ^2 \theta}}$$

by using $$2 \sin \theta \cos \theta=\sin (2\theta)$$

$$\dfrac{a^2}{\cos^2\theta} + \dfrac{b^2}{\sin^2\theta} \ge 2 \sqrt { \dfrac{4a²b²}{\sin^2 (2\theta)}}$$

But now i am totally confused how to deal it to find minimum value of function

Please consider case when $$a>b$$, $$b>a$$

Answer in the book is:- $$(a+b)^2$$

• hang on, do you need the maximum or the minimum value? Please clarify with an edit
– D S
Commented Mar 10, 2023 at 14:42
• Minimum value! brother... And I mentioned it 2 times in the question.. if still needed i can edit! Commented Mar 10, 2023 at 14:47
• the title says the contrary
– D S
Commented Mar 10, 2023 at 14:47
• I'm voting to close this question because the OP themselves aren't aware about the title they wrote Commented Mar 10, 2023 at 14:47
• Hey brother , it is so difficult to me to write this answer beacuse I am not much aware with latex features and phone's keyboard problems and etc. Please edit at your own, if needed Commented Mar 10, 2023 at 14:51

## 1 Answer

$$\dfrac{a^2}{\cos^2\theta} + \dfrac{b^2}{\sin^2\theta}=a^2\tan^2\theta+b^2\cot^2\theta+a^2+b^2$$ Now by $$AM\ge GM$$ we see that the minimum of $$a^2\tan^2\theta+b^2\cot^2\theta$$ is $$2ab$$ $$\implies \dfrac{a^2}{\cos^2\theta} + \dfrac{b^2}{\sin^2\theta}=a^2\tan^2\theta+b^2\cot^2\theta+a^2+b^2\ge 2ab+a^2+b^2=(a+b)^2$$

# Alternative

By Cauchy Schwarz $$\dfrac{a^2}{\cos^2\theta} + \dfrac{b^2}{\sin^2\theta}\ge\frac{(a+b)^2}{\cos^2\theta+\sin^2\theta}=(a+b)^2$$

• Hey amazing solution exactly what I was thinking to have... But I want to know how thought came in your mind to take LCM and use (1= sin²θ + cos²θ) to make it in tan²θ and cot²θ ... I was thinking the same because I can apply am gm inequality on a²tan²θ+b²cot²θ... But just stuck into it... Commented Mar 10, 2023 at 17:31
• @OpenLearner I first rewrote the expression in sec and cosec then applied 1+tan^2x=sec^2x and same for the cosec one Commented Mar 10, 2023 at 18:09