Question

If $\cot \theta =-\dfrac{3}{4}$ and $\theta $ is not in the second quadrant, then $5\sin \theta +10\cos \theta +9\sec \theta +16\csc \theta -4\cot \theta $

Answer 1

Hint: In this problem we need to calculate the value of the given equation. We are given the value of $\cot \theta $ and some information related to the position of the $\theta $. First, we will observe the value of given $\cot \theta $ and list all the possible positions for the $\theta $. Out of these we will find the position of the $\theta $ by using the given information. After that we will list properties of all trigonometric ratios for which $\theta $ lies in a particular quadrant. Now we will calculate the values of all trigonometric ratios by using the trigonometric formulas ${{\csc }^{2}}x-{{\cot }^{2}}x=1$, $\tan x=\dfrac{1}{\cot x}$, ${{\sec }^{2}}x-{{\tan }^{2}}x=1$, $\sin x=\dfrac{1}{\csc x}$, $\cos x=\dfrac{1}{\sec x}$. By using all these formulas, we will calculate the trigonometric values of each ratio and use them in the given equation and simplify the equation by using basic mathematical operations to get the required result.

Complete step by step solution:
Given that, $\cot \theta =-\dfrac{3}{4}$.
From the above equation, we can say that we have given a negative value for $\cot \theta $. In trigonometry we have the negative value of $\cot \theta $ for $\theta $ belongs to the second or fourth quadrant.
In the problem we have given that the $\theta $ does not belong to the second quadrant. So, the $\theta $ must lie in the fourth quadrant.
In the fourth quadrant we have the value of all trigonometric ratios except $\cos $, $\sec $ as negative and the values of $\cos $, $\sec $ are positive.
Finding the value of $\csc \theta $ by using the trigonometric identity ${{\csc }^{2}}x-{{\cot }^{2}}x=1$, then we will get
$\begin{align}
  & {{\csc }^{2}}\theta -{{\left( -\dfrac{3}{4} \right)}^{2}}=1 \\
 & \Rightarrow {{\csc }^{2}}\theta -\dfrac{9}{16}=1 \\
 & \Rightarrow {{\csc }^{2}}\theta =1+\dfrac{9}{16} \\
 & \Rightarrow {{\csc }^{2}}\theta =\dfrac{25}{16} \\
 & \Rightarrow \csc \theta =\pm \dfrac{5}{4} \\
\end{align}$
We have the value of $\csc $ as negative for $\theta $ lies in the fourth quadrant.
So, the value of $\csc \theta $ as $\csc \theta =-\dfrac{5}{4}$.
Finding the value of $\tan \theta $ by using the formula $\tan x=\dfrac{1}{\cot x}$, then we will get
$\begin{align}
  & \tan \theta =\dfrac{1}{-\dfrac{3}{4}} \\
 & \Rightarrow \tan \theta =-\dfrac{4}{3} \\
\end{align}$
Finding the value of $\sec \theta $ by using the formula ${{\sec }^{2}}x-{{\tan }^{2}}x=1$, then we will have
$\begin{align}
  & {{\sec }^{2}}\theta -{{\left( -\dfrac{4}{3} \right)}^{2}}=1 \\
 & \Rightarrow {{\sec }^{2}}\theta -\dfrac{16}{9}=1 \\
 & \Rightarrow {{\sec }^{2}}\theta =1+\dfrac{16}{9} \\
 & \Rightarrow {{\sec }^{2}}\theta =\dfrac{25}{9} \\
 & \Rightarrow \sec \theta =\pm \dfrac{5}{3} \\
\end{align}$
In the fourth quadrant the value of $\sec \theta $ is positive.
So, we have $\sec \theta =\dfrac{5}{3}$.
Finding the value of $\sin \theta $ by using the formula $\sin x=\dfrac{1}{\csc x}$. We have the value of $\csc \theta $ as $-\dfrac{5}{4}$, then we will get
$\begin{align}
  & \sin \theta =\dfrac{1}{-\dfrac{5}{4}} \\
 & \Rightarrow \sin \theta =-\dfrac{4}{5} \\
\end{align}$
Finding the value of $\cos \theta $ by using the formula $\cos x=\dfrac{1}{\sec x}$. We have the value $\sec \theta =\dfrac{5}{3}$, then we will get
$\begin{align}
  & \cos \theta =\dfrac{1}{\dfrac{5}{3}} \\
 & \Rightarrow \cos \theta =\dfrac{3}{5} \\
\end{align}$
We have values of all the trigonometric ratios as
$\sin \theta =-\dfrac{4}{5}$, $\cos \theta =\dfrac{3}{5}$, $\tan \theta =-\dfrac{4}{3}$, $\cot \theta =-\dfrac{3}{4}$, $\csc \theta =-\dfrac{5}{4}$, $\sec \theta =\dfrac{5}{3}$.
From all the above values, the value of $5\sin \theta +10\cos \theta +9\sec \theta +16\csc \theta -4\cot \theta $ will be
$I=5\left( -\dfrac{4}{5} \right)+10\left( \dfrac{3}{5} \right)+9\left( \dfrac{5}{3} \right)+16\left( -\dfrac{5}{4} \right)-4\left( -\dfrac{3}{4} \right)$
Performing basic mathematical operations in the above equation, then we will get
$\begin{align}
  & I=-4+2\left( 3 \right)+3\left( 5 \right)+4\left( -5 \right)+3 \\
 & \Rightarrow I=-1+6+15-20 \\
 & \Rightarrow I=0 \\
\end{align}$

Hence the value of $5\sin \theta +10\cos \theta +9\sec \theta +16\csc \theta -4\cot \theta $ is $0$.

Note: To solve this problem we can also use another method. In this method we will use basic definitions of the trigonometric ratios and construct a triangle with the values we have. After that we have used Pythagoras theorem to find the missing values of the triangle. Now we will again use basic definitions of trigonometric ratios to get the values of all trigonometric ratios. After having all the values, we can simply calculate the required value.