Answer
425.1k+ views
Hint:Use the values of the cotangent function to solve the inner function first.Break the inside angle into two parts keeping in mind the signs of cotangent function in all the quadrants and simplify it to get the required answer.
Formula used:
Complete step-by-step answer:
According to the question, we need to evaluate ${\cot ^{ - 1}}\left( {\cot \dfrac{{5\pi }}{4}} \right)$ or ${\cot ^{ - 1}}\left( {\cot \left( {\pi + \dfrac{\pi }{4}} \right)} \right)$
We know that in a two-dimensional Cartesian plane, cotangents are positive in the first and third quadrants. And clearly, the angle $\left( {\pi + \dfrac{\pi }{4}} \right)$ is lying in the third quadrant, therefore making it easier to solve.
Since angle lie in the third quadrant $ \Rightarrow \cot \left( {\pi + x} \right) = \cot x$
So, the above question can be represented by;
$ \Rightarrow {\cot ^{ - 1}}\left( {\cot \dfrac{{5\pi }}{4}} \right) = {\cot ^{ - 1}}\left( {\cot \left( {\pi + \dfrac{\pi }{4}} \right)} \right) = {\cot ^{ - 1}}\left( {\cot \dfrac{\pi }{4}} \right)$
And also we know that, $\cot \dfrac{\pi }{4} = 1$ , so we get
$ \Rightarrow {\cot ^{ - 1}}\left( {\cot \dfrac{{5\pi }}{4}} \right) = {\cot ^{ - 1}}\left( {\cot \left( {\pi + \dfrac{\pi }{4}} \right)} \right) = {\cot ^{ - 1}}\left( {\cot \dfrac{\pi }{4}} \right) = {\cot ^{ - 1}}1$
Now, we have to solve this inverse trigonometric function. This is basically taking values and returns an angle in radians as output.
But,${\cot ^{ - 1}}1 = \dfrac{\pi }{4} \text{or} \dfrac{{5\pi }}{4}$, because cotangent function obtains $1$ at both $45^\circ$ and $225^\circ $
${\cot ^{ - 1}}\left( {\cot \dfrac{{5\pi }}{4}} \right) = \dfrac{\pi }{4}$
So, the correct answer is “Option A”.
Additional Information:For your information, $\pi $ radian angle is equal to $180^\circ $ and therefore $\dfrac{\pi }{4} = 45^\circ $.Don’t get worried about these inverse functions. Inverse trigonometric functions are simply defined as the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant, and cosecant functions. They are also termed as arcus functions, anti trigonometric functions or cyclometric functions.
Note:Be careful with the angles in radian and degree. Try to go step by step from inside and keep the range and domain in mind. An alternative approach can be by the use of inverse trigonometric functions and solving the problem from outside.
Formula used:
Complete step-by-step answer:
According to the question, we need to evaluate ${\cot ^{ - 1}}\left( {\cot \dfrac{{5\pi }}{4}} \right)$ or ${\cot ^{ - 1}}\left( {\cot \left( {\pi + \dfrac{\pi }{4}} \right)} \right)$
We know that in a two-dimensional Cartesian plane, cotangents are positive in the first and third quadrants. And clearly, the angle $\left( {\pi + \dfrac{\pi }{4}} \right)$ is lying in the third quadrant, therefore making it easier to solve.
Since angle lie in the third quadrant $ \Rightarrow \cot \left( {\pi + x} \right) = \cot x$
So, the above question can be represented by;
$ \Rightarrow {\cot ^{ - 1}}\left( {\cot \dfrac{{5\pi }}{4}} \right) = {\cot ^{ - 1}}\left( {\cot \left( {\pi + \dfrac{\pi }{4}} \right)} \right) = {\cot ^{ - 1}}\left( {\cot \dfrac{\pi }{4}} \right)$
And also we know that, $\cot \dfrac{\pi }{4} = 1$ , so we get
$ \Rightarrow {\cot ^{ - 1}}\left( {\cot \dfrac{{5\pi }}{4}} \right) = {\cot ^{ - 1}}\left( {\cot \left( {\pi + \dfrac{\pi }{4}} \right)} \right) = {\cot ^{ - 1}}\left( {\cot \dfrac{\pi }{4}} \right) = {\cot ^{ - 1}}1$
Now, we have to solve this inverse trigonometric function. This is basically taking values and returns an angle in radians as output.
But,${\cot ^{ - 1}}1 = \dfrac{\pi }{4} \text{or} \dfrac{{5\pi }}{4}$, because cotangent function obtains $1$ at both $45^\circ$ and $225^\circ $
${\cot ^{ - 1}}\left( {\cot \dfrac{{5\pi }}{4}} \right) = \dfrac{\pi }{4}$
So, the correct answer is “Option A”.
Additional Information:For your information, $\pi $ radian angle is equal to $180^\circ $ and therefore $\dfrac{\pi }{4} = 45^\circ $.Don’t get worried about these inverse functions. Inverse trigonometric functions are simply defined as the inverse functions of the basic trigonometric functions which are sine, cosine, tangent, cotangent, secant, and cosecant functions. They are also termed as arcus functions, anti trigonometric functions or cyclometric functions.
Note:Be careful with the angles in radian and degree. Try to go step by step from inside and keep the range and domain in mind. An alternative approach can be by the use of inverse trigonometric functions and solving the problem from outside.
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