Question

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

Answer 1

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

Complete step-by-step answer:
Here we are given to find the maximum value of $\dfrac{1}{{\sec \theta }}$ if $0 < \theta < 90^\circ $ so now first of all we must know that $\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 $\sec \theta = \dfrac{{{\text{hypotenuse}}}}{{{\text{base}}}}$ 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 < \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 $\cos \theta = \dfrac{1}{{\sec \theta }}$ and we know that $\cos \theta $ lies between $ - 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 $0 \leqslant \cos \theta \leqslant 1$
As we know that $\cos \theta = \dfrac{1}{{\sec \theta }}$
Hence $\dfrac{1}{{\sec \theta }}$ lies in the interval $[0,1]$
So the maximum value of $\dfrac{1}{{\sec \theta }}$ is $1$ as we know that for $x = 0^\circ ,\sec \theta $$ = 1$ that is maximum. Hence its maximum value is $1$.

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