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Question: What is the maximum value of \(\dfrac{1}{{\sec \theta }}\) if \(0 < \theta < 90^\circ \)?...

What is the maximum value of 1sec⁡θ\dfrac{1}{{\sec \theta }} if 0<θ<90∘0 < \theta < 90^\circ ?

Explanation

Solution

As we know that cos⁡θ=1sec⁡θ\cos \theta = \dfrac{1}{{\sec \theta }} and the range of cos⁡θ\cos \theta lies between −1 and 1 - 1{\text{ and 1}} for any value of θ\theta then we can find the maximum value of cos⁡θ\cos \theta and that will be equal to the maximum value of 1sec⁡θ\dfrac{1}{{\sec \theta }}.

Complete step-by-step answer:
Here we are given to find the maximum value of 1sec⁡θ\dfrac{1}{{\sec \theta }} if 0<θ<90∘0 < \theta < 90^\circ so now first of all we must know that sec⁡θ\sec \theta is known as the secant function which is one of the trigonometric function which is equal to the ratio of the hypotenuse and base in the right angled triangle. Basically hypotenuse is that side which is larger than the other two sides in the right angled triangle.
So we can say that sec⁡θ=hypotenusebase\sec \theta = \dfrac{{{\text{hypotenuse}}}}{{{\text{base}}}} and this is the trigonometric function whose value is always greater than or equal to one.
So here we can say that denominator is always greater than or equal to one so we can say that the given function will always be less than or equal to one.
As it is given that 0<θ<90∘0 < \theta < 90^\circ therefore we can say that the angle lies in the first quadrant and hence here we know that all the trigonometric functions are always positive and hence here this function will also be positive.
We know that cos⁡θ=1sec⁡θ\cos \theta = \dfrac{1}{{\sec \theta }} and we know that cos⁡θ\cos \theta lies between −1 and 1 - 1{\text{ and 1}} but here we are given that the angle is in the first quadrant hence all are positive or equal to zero so we can say that in the first quadrant 0⩽cos⁡θ⩽10 \leqslant \cos \theta \leqslant 1
As we know that cos⁡θ=1sec⁡θ\cos \theta = \dfrac{1}{{\sec \theta }}
Hence 1sec⁡θ\dfrac{1}{{\sec \theta }} lies in the interval [0,1][0,1]
So the maximum value of 1sec⁡θ\dfrac{1}{{\sec \theta }} is 11 as we know that for x=0∘,sec⁡θx = 0^\circ ,\sec \theta =1 = 1 that is maximum. Hence its maximum value is 11.

Note: We know that sec⁡2θ=1+tan⁡2θ,cosec2θ=1+cot⁡2θ,sin⁡2θ+cos⁡2θ=1{\sec ^2}\theta = 1 + {\tan ^2}\theta ,{\text{cose}}{{\text{c}}^2}\theta = 1 + {\cot ^2}\theta ,{\sin ^2}\theta + {\cos ^2}\theta = 1 and cos⁡θ=1sec⁡θ,tan⁡θ=1cot⁡θ,sin⁡θ=1cosecθ\cos \theta = \dfrac{1}{{\sec \theta }},\tan \theta = \dfrac{1}{{\cot \theta }},\sin \theta = \dfrac{1}{{{\text{cosec}}\theta }} so here we can use these types of formulae.