Prove that:-)
Sec^2 theta + Cos^2 theta will be never less than 2. (WITHOUT USING HIT AND TRIAL METHOD)
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Hey, there!
Here's your answer.
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
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.
Here's your answer.
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
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.
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