Question

How do you find the exact value of \[arctan\left( 2 \right)+arctan\left( 3 \right)\]?

Answer 1

Hint: Arctan is the inverse of the tangent function. Since Tangent is a periodic trigonometric function it doesn’t have an inverse function. But we can have an inverse function of tangent if the function is monotonic in a certain interval. \[arc\tan (x)\] is used to find the inverse value of the tangent function. It can also be written as \[{{\tan }^{-1}}x\].

Complete step by step answer:
As per the given question, we have to find the value of the given expression or function. Here, we have \[arctan\left( 2 \right)+arctan\left( 3 \right)\].
Let us assume that \[\arctan 2+\arctan 3=\alpha \]. As we have the inverse tangent function in the given equation, we take tangents on both sides of the equation. So, when we apply tangent on both sides, we get
\[\Rightarrow \]\[\tan (\arctan 2+\arctan 3)=\tan \alpha \]
We know the formula for tangent of summation of two angles which is given by \[\tan (a+b)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}\], Then, we can expand the left-hand side of the equation as
\[\Rightarrow \]\[\tan (\arctan 2+\arctan 3)=\dfrac{\tan (\arctan 2)+\tan (\arctan 3)}{1-\tan (\arctan 2)\tan (\arctan 3)}\]
We know that \[\tan (\arctan x)=x\]. Using this formula, the above equation becomes
\[\Rightarrow \]\[\dfrac{\tan (\arctan 2)+\tan (\arctan 3)}{1-\tan (\arctan 2)\tan (\arctan 3)}=\tan \alpha \]
\[\Rightarrow \]\[\dfrac{2+3}{1-2\cdot 3}=\tan \alpha \]
\[\Rightarrow \dfrac{5}{1-6}=\tan \alpha \]
\[\Rightarrow \]\[\tan \alpha =-1\]
Since arctan domain is restricted to an interval \[\dfrac{-\pi }{2}\text{ to }\dfrac{\pi }{2}\], we call this interval as the principal interval. So, we write the general solution of \[\alpha \] equal to \[\dfrac{3\pi }{4}+k\pi \].
If we consider the principal interval then \[\alpha \] will be equal to \[\dfrac{-\pi }{4}\].
\[\therefore \] The exact value of \[arctan\left( 2 \right)+arctan\left( 3 \right)\] is \[\dfrac{-\pi }{4}\].

Note:
While solving the inverse trigonometric equations and trigonometric equations, we have to be very careful about the angles satisfying the required conditions. We need to remember that the general solution must give all the solutions. We need to convert the angles properly. we need to check the interval in which the value is satisfying.