Solveeit Logo

Question

Question: Find the principal value of : \({{\cot }^{-1}}\left( \tan \dfrac{3\pi }{4} \right)\)....

Find the principal value of : cot1(tan3π4){{\cot }^{-1}}\left( \tan \dfrac{3\pi }{4} \right).

Explanation

Solution

Hint: The question is related to inverse trigonometric functions. Assume the given function to be equal to xx. Find the value of cotx\cot x. Then find the value of xx 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 cot1(tan3π4){{\cot }^{-1}}\left( \tan \dfrac{3\pi }{4} \right). Let us assume the value of the inverse trigonometric function to be equal to xx. So, we get:
cot1(tan3π4)=x{{\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(cot1(tan3π4))=cotx\cot \left( {{\cot }^{-1}}\left( \tan \dfrac{3\pi }{4} \right) \right)=\cot x
Now, we know the value of cot(cot1y)\cot \left( {{\cot }^{-1}}y \right) is equal to yy. So, we get:
tan3π4=cotx.....(i)\tan \dfrac{3\pi }{4}=\cot x.....(i).
Now, we know, tangent function is negative in the second quadrant. So, the value of tan3π4\tan \dfrac{3\pi }{4} is equal to 1-1 . We will substitute the value of tan3π4\tan \dfrac{3\pi }{4} as 1-1 in equation (i)(i). On substituting the value of tan3π4\tan \dfrac{3\pi }{4} as 1-1 in equation (i)(i), we get:
cotx=1\cot x=-1.
We know, the range for principal value is (0,π)\left( 0,\pi \right). So, we have to find a value of xx such that x(0,π)x\in \left( 0,\pi \right) and cotx=1\cot x=-1. The only possible value which satisfies both conditions is x=3π4x=\dfrac{3\pi }{4}.
So , the value of principal value of the inverse trigonometric function cot1(tan3π4){{\cot }^{-1}}\left( \tan \dfrac{3\pi }{4} \right) is equal to 3π4\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. (0,π)\left( 0,\pi \right)for cotcot function. Students generally forget this condition and end up getting a wrong answer. So, this condition must be satisfied by the obtained principal value.