Find the principal value of : ${{\cot }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)$.
Answer
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Hint: The question is related to inverse trigonometric functions. Assume the given function to be equal to $x$. Find the value of $\cot x$. Then find the value of $x$ which gives the acquired value on applying cotangent function.
Complete step-by-step answer:
We are asked to find the principal value of the inverse trigonometric function ${{\cot }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)$. Let us assume the value of the inverse trigonometric function to be equal to $x$. So, we get:
${{\cot }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)=x$
Now, we will apply cotangent function on both sides of the equation. On applying cotangent function on both sides of the equation, we get:
\[\cot \left( {{\cot }^{-1}}\left( \tan \dfrac{3\pi }{4} \right) \right)=\cot x\]
Now, we know the value of $\cot \left( {{\cot }^{-1}}y \right)$ is equal to $y$. So, we get:
$\tan \dfrac{3\pi }{4}=\cot x.....(i)$.
Now, we know, tangent function is negative in the second quadrant. So, the value of $\tan \dfrac{3\pi }{4}$ is equal to $-1$ . We will substitute the value of $\tan \dfrac{3\pi }{4}$ as $-1$ in equation $(i)$. On substituting the value of $\tan \dfrac{3\pi }{4}$ as $-1$ in equation $(i)$, we get:
$\cot x=-1$.
We know, the range for principal value is $\left( 0,\pi \right)$. So, we have to find a value of $x$ such that $x\in \left( 0,\pi \right)$ and $\cot x=-1$. The only possible value which satisfies both conditions is $x=\dfrac{3\pi }{4}$.
So , the value of principal value of the inverse trigonometric function ${{\cot }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)$ is equal to $\dfrac{3\pi }{4}$.
Note: While solving the problem, make sure that the value of the inverse trigonometric function lies in the principal value range, i.e. $\left( 0,\pi \right)$for \[cot\] function. Students generally forget this condition and end up getting a wrong answer. So, this condition must be satisfied by the obtained principal value.
Complete step-by-step answer:
We are asked to find the principal value of the inverse trigonometric function ${{\cot }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)$. Let us assume the value of the inverse trigonometric function to be equal to $x$. So, we get:
${{\cot }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)=x$
Now, we will apply cotangent function on both sides of the equation. On applying cotangent function on both sides of the equation, we get:
\[\cot \left( {{\cot }^{-1}}\left( \tan \dfrac{3\pi }{4} \right) \right)=\cot x\]
Now, we know the value of $\cot \left( {{\cot }^{-1}}y \right)$ is equal to $y$. So, we get:
$\tan \dfrac{3\pi }{4}=\cot x.....(i)$.
Now, we know, tangent function is negative in the second quadrant. So, the value of $\tan \dfrac{3\pi }{4}$ is equal to $-1$ . We will substitute the value of $\tan \dfrac{3\pi }{4}$ as $-1$ in equation $(i)$. On substituting the value of $\tan \dfrac{3\pi }{4}$ as $-1$ in equation $(i)$, we get:
$\cot x=-1$.
We know, the range for principal value is $\left( 0,\pi \right)$. So, we have to find a value of $x$ such that $x\in \left( 0,\pi \right)$ and $\cot x=-1$. The only possible value which satisfies both conditions is $x=\dfrac{3\pi }{4}$.
So , the value of principal value of the inverse trigonometric function ${{\cot }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)$ is equal to $\dfrac{3\pi }{4}$.
Note: While solving the problem, make sure that the value of the inverse trigonometric function lies in the principal value range, i.e. $\left( 0,\pi \right)$for \[cot\] function. Students generally forget this condition and end up getting a wrong answer. So, this condition must be satisfied by the obtained principal value.
Last updated date: 15th Sep 2023
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