Question

The minimum value of $64\sec \theta +27\operatorname{cosec}\theta$ when $\theta$ lies in $\left( 0,\dfrac{\pi }{2} \right)$ is $A.125$
B.625$C.25$
D. 1025$$$$

Answer 1

We find the critical points $\theta ={{\theta }_{c}}$by equating the derivative of the given function $f\left( \theta \right)=64\sec \theta +27\operatorname{cosec}\theta$ to zero in the form of $\tan \theta$. We check whether the function is minimum by checking the double derivative at $\theta ={{\theta }_{c}}$ is greater than zero or not. We use Pythagorean trigonometric identity ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta ,\cos e{{c}^{2}}\theta =1+{{\cot }^{2}}\theta$ to get the minimum value. $$$$

Complete step-by-step answer:
We are given a trigonometric function in $\theta$ from the question as
$f\left( \theta \right)=64\sec \theta +27\operatorname{cosec}\theta .......\left( 1 \right)$
We are also given that $\theta$ lies in the interval $\left( 0,\dfrac{\pi }{2} \right)$ which means in the first quadrant. Let us differentiate the given function with respect to $\theta$ in order to find the critical points. We have
$\dfrac{d}{d\theta }f\left( \theta \right)=\dfrac{d}{d\theta }\left( 64\sec \theta +27\operatorname{cosec}\theta \right)$
We use rule of sum for differentiation and have;

& \Rightarrow \dfrac{d}{d\theta }f\left( \theta \right)=\dfrac{d}{d\theta }64\sec \theta +\dfrac{d}{d\theta }27\operatorname{cosec}\theta \\\ & \Rightarrow \dfrac{d}{d\theta }f\left( \theta \right)=64\dfrac{d}{d\theta }\sec \theta +27\dfrac{d}{d\theta }\operatorname{cosec}\theta \\\ \end{aligned}$$ We use the standard differentiation formula for $\sec \theta $ and $\operatorname{cosec}\theta $ to have; $$\begin{aligned} & \Rightarrow \dfrac{d}{d\theta }f\left( \theta \right)=64\sec \theta \tan \theta +27\left( -\operatorname{cosec}\theta \cot \theta \right) \\\ & \Rightarrow \dfrac{d}{d\theta }f\left( \theta \right)=64\sec \theta \tan \theta -27\operatorname{cosec}\theta \cot \theta ......\left( 2 \right) \\\ \end{aligned}$$ We equate the above differentiate to zero to have; $$\Rightarrow 64\sec \theta \tan \theta -27\operatorname{cosec}\theta \cot \theta =0$$ We convert all the above trigonometric functions into sin and cosine to have; $$\begin{aligned} & \Rightarrow 64\dfrac{1}{\cos \theta }\dfrac{\sin \theta }{\cos \theta }-27\dfrac{1}{\sin \theta }\dfrac{\cos \theta }{\sin \theta }=0 \\\ & \Rightarrow 64\dfrac{1}{\cos \theta }\dfrac{\sin \theta }{\cos \theta }=27\dfrac{1}{\sin \theta }\dfrac{\cos \theta }{\sin \theta } \\\ \end{aligned}$$ We cross multiply to have; $$\begin{aligned} & \Rightarrow \dfrac{{{\sin }^{3}}\theta }{{{\cos }^{3}}\theta }=\dfrac{27}{64} \\\ & \Rightarrow {{\left( \dfrac{\sin \theta }{\cos \theta } \right)}^{3}}={{\left( \dfrac{3}{4} \right)}^{3}} \\\ \end{aligned}$$ We put back $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ to have; $$\Rightarrow {{\left( \tan \theta \right)}^{3}}={{\left( \dfrac{3}{4} \right)}^{3}}$$ We take cube root both side of the above equation to have; $$\Rightarrow \tan \theta =\dfrac{3}{4}$$ The values of $\theta $ for which the above equation is satisfied are critical points. We know that $\tan \theta $ is a strictly increasing function with respect to $\theta $. Since for values of critical point ${{\theta }_{c}}$ in the first quadrant $\sec \theta ,\operatorname{cosec}\theta$ are positive then the given function $f\left( \theta \right)=64\sec \theta +27\operatorname{cosec}\theta $ is not going to change sign. So we proceed to find the second derivative at $\tan {{\theta }_{c}}=\dfrac{3}{4}$. We have; $$\begin{aligned} & \dfrac{{{d}^{2}}}{d{{\theta }^{2}}}f\left( \theta \right)=64\left( \sec \theta \cdot {{\sec }^{2}}\theta +\tan \theta \cdot \sec \theta \tan \theta \right) \\\ & -27\left( \operatorname{cosec}\theta \cdot \left( -{{\operatorname{cosec}}^{2}}\theta \right)+\operatorname{cosec}\theta \cdot \left( -\operatorname{cosec}\theta \cdot \cot \theta \right) \right) \\\ & \Rightarrow \dfrac{{{d}^{2}}}{d{{\theta }^{2}}}f\left( \theta \right)=64\left( {{\sec }^{3}}\theta +\sec \theta {{\tan }^{2}}\theta \right)+27\left( {{\operatorname{cosec}}^{3}}\theta +{{\operatorname{cosec}}^{2}}\theta \cot \theta \right).....\left( 3 \right) \\\ \end{aligned}$$ We need to double derivative value at the critical point $\tan {{\theta }_{c}}=\dfrac{3}{4}$.We use the Pythagorean identity of secant and tangent to have; $$\sec \theta =\sqrt{1+{{\tan }^{2}}\theta }=\sqrt{1+{{\left( \dfrac{3}{4} \right)}^{2}}}=\dfrac{5}{4}$$ We use the reciprocal relationship between tangent and cotangent to have; $$\cot \theta =\dfrac{1}{\tan \theta }=\dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}$$ We use the Pythagorean identity of cosecant and cotangent to have; $$\operatorname{cosec}\theta =\sqrt{1+{{\cot }^{2}}\theta }=\sqrt{1+{{\left( \dfrac{4}{3} \right)}^{2}}}=\dfrac{5}{3}$$ We put the values $\operatorname{cosec}\theta ,\cot \theta ,\sec \theta ,\tan \theta $ in the double derivative (3) to have; $$\dfrac{{{d}^{2}}}{d{{\theta }^{2}}}f\left( {{\theta }_{c}} \right)=64\left( {{\left( \dfrac{5}{4} \right)}^{3}}+\dfrac{5}{4}\times {{\left( \dfrac{3}{4} \right)}^{2}} \right)+27\left( {{\left( \dfrac{5}{3} \right)}^{3}}+{{\left( \dfrac{5}{4} \right)}^{2}}\left( \dfrac{4}{3} \right) \right)>0.$$ So there is a minimum at $\tan \theta =\dfrac{3}{4}$ for the function $f\left( \theta \right)$ and the minimum value is; $$f\left( \theta \right)=64\sec \theta +27\operatorname{cosec}\theta =64\times \dfrac{5}{4}+27\times \dfrac{5}{3}=125$$ **So, the correct answer is “Option A”.** **Note:** We see that since $\tan \theta $ is a strictly increasing function $\theta ={{\tan }^{-1}}\left( \dfrac{3}{4} \right)$ will be the only critical point in $\left( 0,\dfrac{\pi }{2} \right)$ and we can find the minimum directly without calculating the double derivative. We note that critical points are also the points where the derivative is not defined for example $\theta =\dfrac{\pi }{2}$ for the function $\tan \theta $.