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

Last updated date: 20th Mar 2023

•

Total views: 310.2k

•

Views today: 4.88k

Answer

Verified

310.2k+ views

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.

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?