 QUESTION

# Calculate the value of following: $\operatorname{arccot} [\tan ( - 37^\circ )]$.

Hint: Recall the range of inverse cotangent functions, which is $(0,\pi )$. Convert the angle in tangent to the interval $(0,\pi )$ in terms of cotangent and then solve it to get the final answer.

Inverse trigonometric functions are also referred to as arcus functions or anti-trigonometric functions.
They are inverse functions of the trigonometric functions that have domains that are duly constrained.
Further, they are particularly inverse functions of sine, cosine, tangent, cotangent, secant, and cosecant functions. They are used to attain an angle from any of the angle’s trigonometric ratios.
The inverse cotangent function has a range of values in the interval $(0,\pi )$. Hence, the final angle should be expressed in this interval.
To convert a tangent function into a cotangent function, we use the following identity.
$\tan x = \cot (90^\circ - x)$
We can use this identity to convert tan 37° in terms of cotangent. Hence, we have as follows:
$\operatorname{arccot} [\tan ( - 37^\circ )] = \operatorname{arccot} [\cot (90^\circ - ( - 37^\circ ))]$
We simplify the above expression to get as follows:
$\operatorname{arccot} [\tan ( - 37^\circ )] = \operatorname{arccot} [\cot (90^\circ + 37^\circ )]$
$\operatorname{arccot} [\tan ( - 37^\circ )] = \operatorname{arccot} [\cot (127^\circ )]$
Now, for any function, if we take the inverse of that function, we get an identity function such that the result lies in the range of the inverse function.
${f^{ - 1}}(f(x)) = x$
Inverse trigonometric functions also behave similarly.
The angle 127° lies in the range of inverse cotangent function, hence, we have:
$\operatorname{arccot} [\tan ( - 37^\circ )] = 127^\circ$
Hence, the value of $\operatorname{arccot} [\tan ( - 37^\circ )]$ is 127°.

Note: You can also use the relation between inverse cotangent and inverse tangent function, that is, $\operatorname{arccot} x = \dfrac{\pi }{2} - \arctan x$. The range of the inverse tangent function is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ and then you can proceed.