Question

How do you prove $\tan^{- 1}\left( \dfrac{1}{3} \right) + \tan^{- 1}\left( \dfrac{1}{5} \right) = \tan^{- 1}\left( \dfrac{4}{7} \right)$ ?

Answer 1

In this question, we need to prove that $\tan^{- 1}\left( \dfrac{1}{3} \right) + \tan^{- 1}\left( \dfrac{1}{5} \right)$ is equal to $\tan^{- 1}\left( \dfrac{4}{7} \right)$ . Sine , cosine and tangent are the basic functions of trigonometry . Tangent function is nothing but a ratio of the opposite side of a right angle to the adjacent of the right angle. Tangent inverse is nothing but it is an inverse of the trigonometric function tangent . There is a direct trigonometric formula in the terms of $\tan^{- 1}$ . With the help of the Trigonometric formula , we can easily prove the given expression.
Formula used :
$\tan^{- 1}x + \tan^{- 1}y = \tan^{- 1}\left( \dfrac{x + y}{1 – xy} \right)$

Complete step-by-step solution:
Given, we need to prove $\tan^{- 1}\left( \dfrac{1}{3} \right) + \tan^{- 1}\left( \dfrac{1}{5} \right) = \tan^{- 1}\left( \dfrac{4}{7} \right)$
Let us consider the left side of the given expression,
$\Rightarrow \ \tan^{- 1}\left( \dfrac{1}{3} \right) + \tan^{- 1}\left( \dfrac{1}{5} \right)$
By applying the formula, here $x=\dfrac{1}{3}$ and $y=\dfrac{1}{5}$
$\Rightarrow \ \ \tan^{- 1}\left( \dfrac{1}{3} \right) + \tan^{- 1}\left( \dfrac{1}{5} \right) = \tan^{- 1}\left( \dfrac{\dfrac{1}{3} + \dfrac{1}{5}}{1 - \left( \dfrac{1}{3} \right)\left( \dfrac{1}{5} \right)} \right)$
On simplifying,
We get,
$= \tan^{- 1}\left( \dfrac{\dfrac{5 + 3}{15}}{1 - \left( \dfrac{1}{15} \right)} \right)$
On further simplifying, We get-
$= \tan^{- 1}\left( \dfrac{\dfrac{8}{15}}{\left( \dfrac{15 – 1}{15} \right)} \right)$
By solving, We get
$= \tan^{- 1}\left( \left( \dfrac{8}{15} \right) \times \left( \dfrac{15}{14} \right) \right)$
On simplifying, We get-
$= \tan^{- 1}\left( \dfrac{4}{7} \right)$
Thus we get the right side of the expression.
Hence we have proved $\tan^{- 1}\left( \dfrac{1}{3} \right) + \tan^{- 1}\left( \dfrac{1}{5} \right) = \tan^{- 1}\left( \dfrac{4}{7} \right)$
Final answer :
We have proved $\tan^{- 1}\left( \dfrac{1}{3} \right) + \tan^{- 1}\left( \dfrac{1}{5} \right) = \tan^{- 1}\left( \dfrac{4}{7} \right)$

Note: The concept used in this problem is trigonometric identities and ratios. Trigonometric identities are nothing but they involve trigonometric functions including variables and constants. The common technique used in this problem is the use of trigonometric functions. With the help of properties of the inverse Trigonometric functions we have proved the given expression. The range of $\tan^{- 1}$ is $\left( - \dfrac{\pi}{2},\dfrac{\pi}{2} \right)$. The value of $\dfrac{\pi}{2}$ is equal to $90$ . Similarly the range of $sin^{- 1}$ is $\left\lbrack - \dfrac{\pi}{2},\dfrac{\pi}{2} \right\rbrack$ and the domain of $sin^{- 1}$ is $\lbrack – 1,1\rbrack$ and also the range of $cos^{- 1}$ is $\lbrack 0,\ \pi\rbrack$ and the domain of $cos^{- 1}$ is $\lbrack – 1,\ 1\rbrack$ .