# ${{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3$ is equal to

A. $0$

B. $\pi $

C. $\frac{\pi }{2}$

D. None of these

Last updated date: 24th Mar 2023

•

Total views: 307.5k

•

Views today: 5.85k

Answer

Verified

307.5k+ views

Hint: So for ${{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3$ use ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right)$. Simplify it in a simple manner. Try it and you will get the answer.

Complete step-by-step answer:

Trigonometry has its roots in the right triangle. And so, the tangent defines one of the relationships in that right triangle.

The relationship that the tangent defines is the ratio of the opposite side to the adjacent side of a particular angle of the right triangle.

The function $\tan x$ is defined for all real numbers $x$ such that$\cos x\ne 0$since tangent is the quotient of sine over cosine. Thus $\tan x$ is undefined for$x=......,-\frac{3\pi }{2}.....,\frac{\pi }{2},\frac{3\pi }{2}......$

In a right triangle, the tangent of an angle is the length of the opposite side divided by the length of the adjacent side.

Its range is all real numbers, that is, for any number $y$, you can always find a number $x$ such that $y=\tan x$. The period $\tan x$ is $\pi $. This is a departure from$\sin x$ and $\cos x$, which have periods of $2\pi $.

The reason is simple: opposite angles on the unit circle (like$\frac{\pi }{4}$ and $\frac{5\pi }{4}$ ) have the same tangent because of the signs of their sines and cosines.

The function $\tan x$is an odd function, which you should be able to verify on your own. Finally, at the values of $x$ at which $\tan x$ is undefined, $\tan x$ has both left and right vertical asymptotes.

The tangent function, along with sine and cosine, is one of the three most common trigonometric functions. In any right triangle, the tangent of an angle is the length of the opposite side divided by the length of the adjacent side . In a formula, it is written simply as '$\tan $'.

So now we know that,

${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right)$

So we know${{\tan }^{-1}}(1)=\frac{\pi }{4}$.

So we have given ${{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3=\frac{\pi }{4}+{{\tan }^{-1}}2+{{\tan }^{-1}}3$

Now applying the above property we get,

$=\frac{\pi }{4}+{{\tan }^{-1}}\left( \frac{2+3}{1-2\times 3} \right)=\frac{\pi }{4}+{{\tan }^{-1}}\left( -1 \right)$

So we know${{\tan }^{-1}}(-1)=\frac{3\pi }{4}$.

So we get,

$=\frac{\pi }{4}+\frac{3\pi }{4}=\pi $

So we have got the final answer ${{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3=\pi $.

So the correct answer is an option(B).

Note: Carefully read the question. So you should know the identities of $\tan x$.

Most of the students make mistakes in substituting the value ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right)$. So avoid the mistakes.

Complete step-by-step answer:

Trigonometry has its roots in the right triangle. And so, the tangent defines one of the relationships in that right triangle.

The relationship that the tangent defines is the ratio of the opposite side to the adjacent side of a particular angle of the right triangle.

The function $\tan x$ is defined for all real numbers $x$ such that$\cos x\ne 0$since tangent is the quotient of sine over cosine. Thus $\tan x$ is undefined for$x=......,-\frac{3\pi }{2}.....,\frac{\pi }{2},\frac{3\pi }{2}......$

In a right triangle, the tangent of an angle is the length of the opposite side divided by the length of the adjacent side.

Its range is all real numbers, that is, for any number $y$, you can always find a number $x$ such that $y=\tan x$. The period $\tan x$ is $\pi $. This is a departure from$\sin x$ and $\cos x$, which have periods of $2\pi $.

The reason is simple: opposite angles on the unit circle (like$\frac{\pi }{4}$ and $\frac{5\pi }{4}$ ) have the same tangent because of the signs of their sines and cosines.

The function $\tan x$is an odd function, which you should be able to verify on your own. Finally, at the values of $x$ at which $\tan x$ is undefined, $\tan x$ has both left and right vertical asymptotes.

The tangent function, along with sine and cosine, is one of the three most common trigonometric functions. In any right triangle, the tangent of an angle is the length of the opposite side divided by the length of the adjacent side . In a formula, it is written simply as '$\tan $'.

So now we know that,

${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right)$

So we know${{\tan }^{-1}}(1)=\frac{\pi }{4}$.

So we have given ${{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3=\frac{\pi }{4}+{{\tan }^{-1}}2+{{\tan }^{-1}}3$

Now applying the above property we get,

$=\frac{\pi }{4}+{{\tan }^{-1}}\left( \frac{2+3}{1-2\times 3} \right)=\frac{\pi }{4}+{{\tan }^{-1}}\left( -1 \right)$

So we know${{\tan }^{-1}}(-1)=\frac{3\pi }{4}$.

So we get,

$=\frac{\pi }{4}+\frac{3\pi }{4}=\pi $

So we have got the final answer ${{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3=\pi $.

So the correct answer is an option(B).

Note: Carefully read the question. So you should know the identities of $\tan x$.

Most of the students make mistakes in substituting the value ${{\tan }^{-1}}x+{{\tan }^{-1}}y={{\tan }^{-1}}\left( \frac{x+y}{1-xy} \right)$. So avoid the mistakes.

Recently Updated Pages

Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main