
The general solution of the equation $\cot \theta -\tan \theta =2$ is
A. \[n\pi +\dfrac{\pi }{4}\]
B. \[\dfrac{n\pi }{2}+\dfrac{\pi }{8}\]
C. \[\dfrac{n\pi }{2}\pm \dfrac{\pi }{8}\]
D. None of these.
Answer
161.4k+ views
Hint: To determine the general value of $\theta $, we will first substitute the value of cot and tan as $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$in the given equation then we will simplify it and derive an equation. Using the double angles formula we will find the value of angle $\theta $ with an integer $n$.
Formula used:
The double-angle formulas are:
$\begin{align}
& \cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\
& \sin 2\theta =2\sin \theta \cos \theta
\end{align}$
Complete step by step Solution:
We are given $\cot \theta -\tan \theta =2$and we have to determine the general solution.
We know that the trigonometric function tan is the ratio of the function sin to the function cos and cot is the ratio of the function cos to the function sin. That is,$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$.
We will now substitute $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$in the given equation$\cot \theta -\tan \theta =2$.\[\begin{align}
& \cot \theta -\tan \theta =2 \\
& \dfrac{\cos \theta }{\sin \theta }-\dfrac{\sin \theta }{\cos \theta }=2
\end{align}\]
We will now simplify the equation.
\[\begin{align}
& \dfrac{\cos \theta }{\sin \theta }-\dfrac{\sin \theta }{\cos \theta }=2 \\
& \dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{\sin \theta \cos \theta }=2 \\
& {{\cos }^{2}}\theta -{{\sin }^{2}}\theta =2\sin \theta \cos \theta
\end{align}\]
We will now use the double angle formulas.
\[\begin{align}
& {{\cos }^{2}}\theta -{{\sin }^{2}}\theta =2\sin \theta \cos \theta \\
& \cos 2\theta =\sin 2\theta \\
& 1=\dfrac{\sin 2\theta }{\cos 2\theta } \\
& 1=\tan 2\theta
\end{align}\]
Now we know that $\tan \dfrac{\pi }{4}=1$. So,
\[\begin{align}
& \tan \dfrac{\pi }{4}=\tan 2\theta \\
& 2\theta =n\pi +\dfrac{\pi }{4} \\
& \theta =\dfrac{n\pi }{2}+\dfrac{\pi }{8}
\end{align}\]
Here the variable\[n\] is an integer such that \[n\in Z\].
\[\theta =\dfrac{n\pi }{2}+\dfrac{\pi }{8}\]
This general equation we derived $\theta =n\pi \pm \dfrac{\pi }{6}$will give all the solutions for the trigonometric function by substituting the value of $n$.
The general value of $\theta $ is \[\theta =\dfrac{n\pi }{2}+\dfrac{\pi }{8}\] when $\cot \theta -\tan \theta =2$
Therefore, the correct option is (B).
Note: The interval of the function tan and cot is $0\le \theta \le \pi $ which means that it repeats its values after an interval of $\pi $. The domain of the trigonometric function tan is the set of all the real numbers except the values for which the function cos will be zero while the domain of cot contains all the real numbers except for those values where the function tan will be zero.
Formula used:
The double-angle formulas are:
$\begin{align}
& \cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\
& \sin 2\theta =2\sin \theta \cos \theta
\end{align}$
Complete step by step Solution:
We are given $\cot \theta -\tan \theta =2$and we have to determine the general solution.
We know that the trigonometric function tan is the ratio of the function sin to the function cos and cot is the ratio of the function cos to the function sin. That is,$\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$.
We will now substitute $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$in the given equation$\cot \theta -\tan \theta =2$.\[\begin{align}
& \cot \theta -\tan \theta =2 \\
& \dfrac{\cos \theta }{\sin \theta }-\dfrac{\sin \theta }{\cos \theta }=2
\end{align}\]
We will now simplify the equation.
\[\begin{align}
& \dfrac{\cos \theta }{\sin \theta }-\dfrac{\sin \theta }{\cos \theta }=2 \\
& \dfrac{{{\cos }^{2}}\theta -{{\sin }^{2}}\theta }{\sin \theta \cos \theta }=2 \\
& {{\cos }^{2}}\theta -{{\sin }^{2}}\theta =2\sin \theta \cos \theta
\end{align}\]
We will now use the double angle formulas.
\[\begin{align}
& {{\cos }^{2}}\theta -{{\sin }^{2}}\theta =2\sin \theta \cos \theta \\
& \cos 2\theta =\sin 2\theta \\
& 1=\dfrac{\sin 2\theta }{\cos 2\theta } \\
& 1=\tan 2\theta
\end{align}\]
Now we know that $\tan \dfrac{\pi }{4}=1$. So,
\[\begin{align}
& \tan \dfrac{\pi }{4}=\tan 2\theta \\
& 2\theta =n\pi +\dfrac{\pi }{4} \\
& \theta =\dfrac{n\pi }{2}+\dfrac{\pi }{8}
\end{align}\]
Here the variable\[n\] is an integer such that \[n\in Z\].
\[\theta =\dfrac{n\pi }{2}+\dfrac{\pi }{8}\]
This general equation we derived $\theta =n\pi \pm \dfrac{\pi }{6}$will give all the solutions for the trigonometric function by substituting the value of $n$.
The general value of $\theta $ is \[\theta =\dfrac{n\pi }{2}+\dfrac{\pi }{8}\] when $\cot \theta -\tan \theta =2$
Therefore, the correct option is (B).
Note: The interval of the function tan and cot is $0\le \theta \le \pi $ which means that it repeats its values after an interval of $\pi $. The domain of the trigonometric function tan is the set of all the real numbers except the values for which the function cos will be zero while the domain of cot contains all the real numbers except for those values where the function tan will be zero.
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