The problem:
Prove that sec^2 theta + cos^2 theta can never be less than 2.
I can't figure out how to go about.
I have tried to take sec^ theta as 1/cos^2 theta and cos^2 theta as sin^2 theta, but got nowhere.
Let us take, $\displaystyle f(\theta)=\sec^2{ \theta} + \cos^2{\theta}$.
Now, use your knowledge of calculus to find the minimum value of $\displaystyle f(\theta)$.
You will find that minimum value of $\displaystyle f(\theta)$ is 2.
Therefore:
$\displaystyle f(\theta) \geq 2$
$\displaystyle \sec^2{ \theta} + \cos^2{\theta} \geq 2$
Thanks for the reply.
I am still hazy about Calculus, but I think that using algebraic identities like (a+b)^2 and (a-b)^2 this can be solved:
sec^2 theta =1/ cos^2 theta. Therefore, cos^2 theta can be taken as 'x' and the equation reduces to x^2 + 1/x^2.
x^2 + 1/x^2 - 2*x*1/x + 2*x*1/x.
(x^2 - 1/x^2)^2 + 2
This way, sec^2 theta + cos^2 theta can never be less than 2. Is this right?
Yup it's correct, even if you didn't write it ordered or... well, logical.
$\displaystyle x^{2}+\frac{1}{x^{2}}> 2 \Leftrightarrow x^{2}+\frac{1}{x^{2}}-2>0 \Leftrightarrow x^{2}+\frac{1}{x^{2}}-2\cdot x^{2}\cdot\frac{1}{x^{2}}>0 \Leftrightarrow \left ( x-\frac{1}{x} \right )^{2}>0$