min. positive value of a^2 tan^2 (theta) + b^2 cot^2(theta)

[defeated_soldier]

I want to calculate the minimum positive value of a^2 tan^2 (theta) + b^2 cot^2(theta)

How do i start ?

 

[stapel]

I want to calculate the minimum positive value.... How do i start ?

How about starting by taking the derivative with respect to theta?

Eliz.

 

[soroban]

Re: min. positive value of a^2 tan^2 (theta) + b^2 cot^2(the

Hello, defeated_soldier!

Find the minimum positive value of: \(\displaystyle \L\,f(x)\:=\:a^2\cdot\tan^2\theta \,+\,b^2\cdot\cot^2\theta\)

Are you new to maximizing/minimizing?

As Eliz.Stapel suggested, find the derivative.
. . Set it equal to 0 and solve.

\(\displaystyle \L f'(x)\;=\;2a^2\cdot\tan\theta\cdot\sec^2\theta\,-\,2b^2\cdot\cot\theta\cdot\csc^2\theta \:=\:0\)


We have: \(\displaystyle \L\:a^2\cdot\tan\theta\cdot\sec^2\theta\:=\:b^2\cdot\cot\theta\cdot\csc^2\theta\)

. . . . \(\displaystyle \L a^2\cdot\frac{\sin\theta}{\cos\theta}\cdot\frac{1}{\cos^2\theta} \;=\;b^2\cdot\frac{\cos\theta}{\sin\theta}\cdot\frac{1}{\sin^2\theta}\;\;\Rightarrow\;\;\frac{\sin^4\theta}{\cos^4\theta} \:=\:\frac{b^2}{a^2}\)

. . . . \(\displaystyle \L\tan^4\theta\:=\:\frac{b^2}{a^2}\;\;\Rightarrow\;\;\tan^2\theta\:=\:\frac{b}{a}\,\) and \(\displaystyle \L\,\cot^2\theta\:=\:\frac{a}{b}\)


Substitute into the function:
. . \(\displaystyle \L f(x)\;=\;a^2\left(\frac{b}{a}\right)\,+\,b^2\left(\frac{a}{b}\right) \;=\;ab\,+\,ba\;=\;2ab\)