Question

How do you write an algebraic expression that is equivalent to$\cot \left( {\arctan x} \right)$?

Answer 1

Hint: In order to determine the algebraic equivalent of given expression, use the property of trigonometry that$\cot \alpha = \dfrac{1}{{\tan \left( \alpha \right)}}$. Now substitute $\alpha $with $\arctan x$and write the denominator of the RHS part equal to $x$as $\tan $and $\,\arctan $are the inverse of each other.

Complete step by step answer:
We are given a trigonometric expression as $\cot \left( {\arctan x} \right)$ and we have to find its algebraic equivalent.
So, to find the algebraic expression, we will use the property of trigonometry that the cotangent is equal to the reciprocal of tangent.
i.e. $\cot \alpha = \dfrac{1}{{\tan \left( \alpha \right)}}$
Putting $\alpha = \arctan x$in the above expression we get
$\cot \left( {\arctan x} \right) = \dfrac{1}{{\tan \left( {\arctan x} \right)}}$
The denominator in the Right-hand side of the equation will become $x$ as $\tan $and $\,\arctan $are basically the inverse of each other and for inverse function we can write :$f\left( {{f^{ - 1}}\left( x \right)} \right) = {f^{ - 1}}\left( {f\left( x \right)} \right) = x$
 We obtain the expression after all transformations as
$\therefore \cot \left( {\arctan x} \right) = \dfrac{1}{x}$

Therefore, the algebraic equivalent of $\cot \left( {\arctan x} \right)$ is equal to $\dfrac{1}{x}$.

Additional information:
1. In Mathematics the inverse trigonometric functions (every so often additionally called anti-trigonometric functions or cyclometric function) are the reverse elements of the mathematical functions In particular, they are the inverses of the sine, cosine, tangent, cotangent, secant, and cosecant functions, and are utilized to get a point from any of the point's mathematical proportions. Reverse trigonometric functions are generally utilized in designing, route, material science, and calculation.
2. In inverse trigonometric function, the domain are the ranges of corresponding trigonometric functions and the range are the domain of the corresponding trigonometric function.
3. Periodic Function= A function $f(x)$ is said to be a periodic function if there exists a real number T > 0 such that $f(x + T) = f(x)$ for all x.
If T is the smallest positive real number such that $f(x + T) = f(x)$ for all x, then T is called the fundamental period of $f(x)$ .
Since $\sin \,(2n\pi + \theta ) = \sin \theta $ for all values of $\theta $ and n$ \in $N.
4. Even Function – A function $f(x)$ is said to be an even function ,if $f( - x) = f(x)$for all x in its domain.
Odd Function – A function $f(x)$ is said to be an even function ,if $f( - x) = - f(x)$for all x in its domain.
We know that $\sin ( - \theta ) = - \sin \theta .\cos ( - \theta ) = \cos \theta \,and\,\tan ( - \theta ) = - \tan \theta $
Therefore,$\sin \theta $ and $\tan \theta $ and their reciprocals,$\cos ec\theta $ and $\cot \theta $ are odd functions whereas \[\cos \theta \] and its reciprocal \[\sec \theta \] are even functions.
5. Trigonometry is one of the significant branches throughout the entire existence of mathematics and this idea is given by a Greek mathematician Hipparchus.

Note: 1.We must be very careful about using the correct identity and property of the trigonometric functions as using inappropriate relations will lead to inaccurate answers.
2.Range and domain of function $\arctan x$is $\left( { - \infty , + \infty } \right)$and $\left( { - \dfrac{\pi }{2}, + \dfrac{\pi }{2}} \right)$ respectively.
3. Avoid step jumps in such type of questions as much as possible because this will increase the chances for silly mistakes.