Question

If $\cos \theta =\dfrac{-3}{5},\pi <\theta <\dfrac{3\pi }{2}$, find the values of other five trigonometric functions and hence evaluate $\dfrac{\csc \theta +\cot \theta }{\sec \theta -\tan \theta }$.

Answer 1

Hint: Use the identity ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ to find the value of $\sin \theta $. Hence find the values of $\sec \theta $ and $\csc \theta $ using the identities $\sec \theta =\dfrac{1}{\cos \theta }$ and $\csc \theta =\dfrac{1}{\sin \theta }$ respectively. Hence find the value of $\tan \theta $ and $\cot \theta $ using the identities $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{1}{\tan \theta }$ respectively. Hence substitute the values of $\csc \theta ,\cot \theta ,\sec \theta $ and $\tan \theta $ to get the value of $\dfrac{\csc \theta +\cot \theta }{\sec \theta -\tan \theta }$.

Complete step-by-step answer:
We have $\cos \theta =\dfrac{-3}{5}$
We know that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$
Substituting the value of $\cos \theta ,$ we get
\[{{\sin }^{2}}\theta +{{\left( \dfrac{-3}{5} \right)}^{2}}=1\Rightarrow {{\sin }^{2}}\theta +\dfrac{9}{25}=1\]
Subtracting $\dfrac{9}{25}$ from both sides, we get
${{\sin }^{2}}\theta =1-\dfrac{9}{25}=\dfrac{16}{25}$
Hence, we have $\sin \theta =\pm \dfrac{4}{5}$
Since $\pi <\theta <\dfrac{3\pi }{2}$, we have $\theta $ lies in the third quadrant.
Since sine of an angle in third quadrant is negative, we have
$\sin \theta =\dfrac{-4}{5}$.
Now, we know that $\csc \theta =\dfrac{1}{\sin \theta }$
Hence, we have $\csc \theta =\dfrac{1}{\dfrac{-4}{5}}=\dfrac{-5}{4}$
Also, $\sec \theta =\dfrac{1}{\cos \theta }$
Hence, we have $\sec \theta =\dfrac{1}{\dfrac{-3}{5}}=\dfrac{-5}{3}$.
We know that $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$
Hence, we have $\tan \theta =\dfrac{\dfrac{-4}{5}}{\dfrac{-3}{5}}=\dfrac{4}{3}$.
Using $\cot \theta =\dfrac{1}{\tan \theta }$, we get
$\cot \theta =\dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}$
Hence, we have $\dfrac{\csc \theta +\cot \theta }{\sec \theta -\tan \theta }=\dfrac{\dfrac{-5}{4}+\dfrac{3}{4}}{\dfrac{-5}{3}-\dfrac{4}{3}}=\dfrac{\dfrac{-5+3}{4}}{\dfrac{-5-4}{3}}=\dfrac{-2}{4}\times \dfrac{3}{-9}=\dfrac{1}{6}$
Hence, $\dfrac{\csc \theta +\cot \theta }{\sec \theta -\tan \theta }=\dfrac{1}{6}$

Note: Alternative solution:
Let ABC be a right-angled triangle right-angled at A. Let AC = -3 units(negative sign only describes the quadrant in which the angle lies) and BC = 5 units. Let $\angle C=\theta $.

Now, we have
$A{{B}^{2}}+A{{C}^{2}}=B{{C}^{2}}\Rightarrow A{{B}^{2}}=25-9=16\Rightarrow AB=\pm 4$
Since $\theta $ lies in the third quadrant, we have $AB<0$
Hence, AB = -4 units.
Now, we have $\sin \theta =\dfrac{AB}{BC}=\dfrac{-4}{5}$
Hence, we have $\sin \theta =\dfrac{-4}{5}$
Similarly, we have $\tan \theta =\dfrac{AB}{AC}=\dfrac{-4}{-3}=\dfrac{4}{3},\cot \theta =\dfrac{AC}{AB}=\dfrac{3}{4},\sec \theta =\dfrac{BC}{AC}=\dfrac{-5}{3},\csc \theta =\dfrac{BC}{AB}=\dfrac{-5}{4}$, which is the same as obtained above.
Proceeding similarly, we can find the value of $\dfrac{\csc \theta +\cot \theta }{\sec \theta -\tan \theta }$.