Question

# What is ${\text{ta}}{{\text{n}}^{ - 1}}(\dfrac{1}{4}){\text{ + ta}}{{\text{n}}^{ - 1}}(\dfrac{3}{5})$ equal to?A. 0B. $\dfrac{\pi }{4}$ C. $\dfrac{\pi }{3}$ D. $\dfrac{\pi }{2}$

Hint:- The inverse trigonometry formulas of ${\text{ta}}{{\text{n}}^{ - 1}}\left( {} \right)$ can be used.

Given,
${\text{ta}}{{\text{n}}^{ - 1}}(\dfrac{1}{4}){\text{ + ta}}{{\text{n}}^{ - 1}}(\dfrac{3}{5})$ =? -(1)
We know that , the inverse trigonometry formula of addition of ${\text{ta}}{{\text{n}}^{ - 1}}\left( {} \right)$ is ${\text{ta}}{{\text{n}}^{ - 1}}{\text{x + ta}}{{\text{n}}^{ - 1}}{\text{y = ta}}{{\text{n}}^{ - 1}}(\dfrac{{{\text{x + y}}}}{{1 - {\text{xy}}}})$ , xy<1 -(2)

Comparing the equation (1) with the equation (2) we get,
X = $\dfrac{1}{4}$ and y =$\dfrac{3}{5}$ .
We need to check whether xy<1 for applying the formula of ${\text{ta}}{{\text{n}}^{ - 1}}\left( {} \right)$
$\Rightarrow {\text{xy = }}\left( {\dfrac{1}{4}} \right)\left( {\dfrac{3}{5}} \right){\text{ = }}\left( {\dfrac{3}{{20}}} \right) \\ \\$
And,
$\Rightarrow \dfrac{3}{{20}} < 1 \\ \Rightarrow {\text{xy < 1}} \\$
So, the formula is applicable for a given set of x and y.

Putting the value of x and y in equation (2). We get,
${\text{ta}}{{\text{n}}^{ - 1}}\dfrac{1}{4}{\text{ + ta}}{{\text{n}}^{ - 1}}\dfrac{3}{5}{\text{ = ta}}{{\text{n}}^{ - 1}}(\dfrac{{\dfrac{1}{4}{\text{ + }}\dfrac{3}{5}}}{{1 - \left( {\dfrac{1}{4}} \right)\left( {\dfrac{3}{5}} \right)}})$

Solving right hand side , we get
${\text{ta}}{{\text{n}}^{ - 1}}\dfrac{1}{4}{\text{ + ta}}{{\text{n}}^{ - 1}}\dfrac{3}{5}{\text{ = ta}}{{\text{n}}^{ - 1}}(\dfrac{{\dfrac{{17}}{{20}}}}{{1 - \left( {\dfrac{3}{{20}}} \right)}})$
${\text{ta}}{{\text{n}}^{ - 1}}\dfrac{1}{4}{\text{ + ta}}{{\text{n}}^{ - 1}}\dfrac{3}{5}{\text{ = ta}}{{\text{n}}^{ - 1}}\left( {\dfrac{{\dfrac{{17}}{{20}}}}{{\dfrac{{17}}{{20}}}}} \right)$
${\text{ta}}{{\text{n}}^{ - 1}}\dfrac{1}{4}{\text{ + ta}}{{\text{n}}^{ - 1}}\dfrac{3}{5}{\text{ = ta}}{{\text{n}}^{ - 1}}1$
${\text{ta}}{{\text{n}}^{ - 1}}\dfrac{1}{4}{\text{ + ta}}{{\text{n}}^{ - 1}}\dfrac{3}{5}{\text{ = }}\dfrac{\pi }{4}$

Hence the value of ${\text{ta}}{{\text{n}}^{ - 1}}(\dfrac{1}{4}){\text{ + ta}}{{\text{n}}^{ - 1}}(\dfrac{3}{5})$ is $\dfrac{\pi }{4}$. The answer is option B.
Note:- The domain and the range of ${\text{ta}}{{\text{n}}^{ - 1}}\left( {} \right)$ is R and $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ respectively. And ${\text{ta}}{{\text{n}}^{ - 1}}{\text{x + ta}}{{\text{n}}^{ - 1}}{\text{y = ta}}{{\text{n}}^{ - 1}}(\dfrac{{{\text{x + y}}}}{{1 - {\text{xy}}}})$ is applicable only when xy <1.