Trigonometry (4)

Geometry Level pending

If x + y = 2 α x+y=2\alpha , x , y ∈ ( 0 , π 2 ) x,y \in \left(0, \frac{\pi}{2} \right) , then the minimum value of sec ⁡ x + sec ⁡ y \sec x+\sec y is:

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cos ⁡ 2 α \cos 2\alpha 2 cos ⁡ α 2\cos \alpha 2 sec ⁡ α 2\sec \alpha None of these.

Sahil Silare by Sahil Silare , 20, India

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1 solution

Md Zuhair
Mar 6, 2017

Here we go, We take cos ⁡ 2 x \cos^2 x and cos ⁡ 2 y \cos^2 y

Now applying AM-GM Inequality

c o s 2 x + c o s 2 y 2 ≥ c o s 2 x c o s 2 y \dfrac{cos^2x+cos^2y}{2} \geq \sqrt{cos^2x cos^2y}

So c o s x ≥ c o s y cosx \geq cos y

Now As x,y is in 1st qudrant

x ≥ y x \geq y

So we can say 2 x ≥ 2 α 2x \geq 2\alpha - or

x ≥ α x \geq \alpha

So minimum occurs at x = y = α x=y=\alpha

Hence m i n ( sec ⁡ x + sec ⁡ y ) {min(\sec x + \sec y)} = 2 sec ⁡ α 2\sec \alpha

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