# What is \[{\text{ta}}{{\text{n}}^{ - 1}}(\dfrac{1}{4}){\text{ + ta}}{{\text{n}}^{ - 1}}(\dfrac{3}{5})\] equal to?

A. 0

B. $\dfrac{\pi }{4}$

C. $\dfrac{\pi }{3}$

D. $\dfrac{\pi }{2}$

Answer

Verified

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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.

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.

Last updated date: 28th Sep 2023

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