Question

Prove that:-)
Sec^2 theta + Cos^2 theta will be never less than 2. (WITHOUT USING HIT AND TRIAL METHOD)

Answer 1

Hey, there!

Here's your answer.

${ \sec }^{2}θ + { \cos }^{2} θ = \frac{1}{ { \cos }^{2}θ } + { \cos}^{2} θ \\ \\ = \frac{1 + { \cos }^{4} θ}{ { \cos }^{2}θ }$

$For \: minimum \: value \: of \: \: \frac{1 + { \cos }^{4} θ}{ { \cos }^{2}θ } \\ \\ \frac{d}{dθ} \: (\frac{1 + { \cos }^{4} θ}{ { \cos }^{2}θ } ) = 0$

Differentiating, we get,

–4cos³θsinθcos²θ + 2sinθcosθ(1 + cos⁴θ) = 0

4cos⁴θ = 2(1 + cos⁴θ)

2cos⁴θ = 1 + cos⁴θ

cos⁴θ = 1

That means, θ = nπ where n ∈ Z

cos(nπ) = ± 1

cos²(nπ) = cos⁴(nπ) = 1

$So, \: minimum \: value \: of \: \frac{1 + { \cos }^{4} θ}{ { \cos }^{2}θ } \: \\ \\ = \frac{1 + { \cos }^{4} n\pi}{ { \cos }^{2}n\pi } \: = \frac{1 + 1}{1} = 2$
Therefore, the minimum value of the function is 2 and maximum value is +∞ (when cosθ = 0)

(See the graph in the attachment)



Hope that helps.

Thanks.