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Question: Prove that \[{\cot ^{ - 1}}( - x) = \pi - {\cot ^{ - 1}}x,x \in R\]...

Prove that cot⁡−1(−x)=π−cot⁡−1x,x∈R{\cot ^{ - 1}}( - x) = \pi - {\cot ^{ - 1}}x,x \in R

Explanation

Solution

Hint: We prove the LHS of the equation equal to the RHS of the equation by considering the LHS of the equation equal to a variable. Apply the function ‘cot’ on both sides so as to cancel the inverse of the function. Use the quadrant diagram of trigonometric functions to convert the required values.

  • Inverse of any function when applied on the same function cancels out the function and the inverse, i.e. f−1(f(x))=x{f^{ - 1}}(f(x)) = x.

Complete step-by-step answer:

We have to prove cot⁡−1(−x)=π−cot⁡−1x,x∈R{\cot ^{ - 1}}( - x) = \pi - {\cot ^{ - 1}}x,x \in R

Consider the left hand side of the equation.

LHS is cot⁡−1(−x){\cot ^{ - 1}}( - x)

We consider the value in LHS as a variable, say ‘y’

⇒cot⁡−1(−x)=y \Rightarrow {\cot ^{ - 1}}( - x) = y

Now apply the function ‘cot’ on both sides of the equation.

\Rightarrow \cot \left\\{ {{{\cot }^{ - 1}}( - x)} \right\\} = \cot (y)

Since, we know f−1(f(x))=x{f^{ - 1}}(f(x)) = x

⇒(−x)=cot⁡(y) \Rightarrow ( - x) = \cot (y) … (1)

Now we look at the quadrant diagram and try to find an angle that gives us the required value in RHS of the equation (1).

Since LHS of equation (1) has a negative value, we try to write the term in RHS as a negative term so as to cancel negative signs from both sides.

From the diagram we know that ‘tan’ and ‘cot’ functions are positive in the first and third quadrant, and negative in the second and fourth quadrant. So, we have cot⁡(π−θ)=−cot⁡θ\cot (\pi - \theta ) = - \cot \theta , for θ\theta at any angle.

So, to change the sign from positive to negative we either write the angle as π−y\pi - yor2π−y2\pi - y.

As we see the RHS in the question has valueπ\pi , so we choose to form the angleπ−y\pi - y.

We can write −cot⁡(y)=cot⁡(π−y) - \cot (y) = \cot (\pi - y)

⇒cot⁡(y)=−cot⁡(π−y) \Rightarrow \cot (y) = - \cot (\pi - y) … (2)

Substitute the value from equation (2) in equation (2)

⇒(−x)=−cot⁡(π−y) \Rightarrow ( - x) = - \cot (\pi - y)

Cancel negative signs from both sides of the equation.

⇒x=cot⁡(π−y) \Rightarrow x = \cot (\pi - y)

Now again apply the inverse ‘cot’ function on both sides of the equation.

\Rightarrow {\cot ^{ - 1}}(x) = {\cot ^{ - 1}}\left\\{ {\cot (\pi - y)} \right\\}

Since, we know f−1(f(x))=x{f^{ - 1}}(f(x)) = x

⇒cot⁡−1(x)=(π−y) \Rightarrow {\cot ^{ - 1}}(x) = (\pi - y)

Shift ‘y’ to one side of the equation, and all other values to the other side of the equation.

⇒y=π−cot⁡−1(x) \Rightarrow y = \pi - {\cot ^{ - 1}}(x)

Substitute the value of cot⁡−1(−x)=y{\cot ^{ - 1}}( - x) = y on LHS of the equation.

⇒cot⁡−1(−x)=π−cot⁡−1(x) \Rightarrow {\cot ^{ - 1}}( - x) = \pi - {\cot ^{ - 1}}(x)

∴\therefore LHS ==RHS

Hence proved.

Note: Students might try to apply inverse function in step where there is negative sign on one side of the equation and positive sign on the other side of the equation, but it is difficult to find the inverse cot of a negative value, so we have to convert the angle using a quadrant diagram.