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# ${{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3$ is equal toA. $0$B. $\pi$C. $\frac{\pi }{2}$D. None of these

Last updated date: 23rd May 2024
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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.

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.