Question

The value of ${{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3$ is
(a) 0
(b) 1
(c) $\pi $
(d) $-\pi $

Answer 1

Hint: Use the inverse trigonometric formula ${{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi +{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ if $xy>1$ and ${{\tan }^{-1}}\left( -x \right)= -{{\tan }^{-1}}x$ to simplify the given expression. Cancel out the common terms to calculate the exact value of the given trigonometric expression.

Complete step by step answer:
We have to calculate the value of ${{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3$. We will simplify the given expression using the trigonometric identity.
We know the trigonometric identity for the inverse of tangent function ${{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi +{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ if $xy>1$.
Substituting $x=1,y=3$ in the above formula, we have ${{\tan }^{-1}}1+{{\tan }^{-1}}3=\pi +{{\tan }^{-1}}\left( \dfrac{1+3}{1-1\times 3} \right)$ as $1\times 3=3>1$.
Simplifying the above expression, we have ${{\tan }^{-1}}1+{{\tan }^{-1}}3=\pi +{{\tan }^{-1}}\left( \dfrac{1+3}{1-1\times 3} \right)=\pi +{{\tan }^{-1}}\left( \dfrac{4}{1-3} \right)=\pi +{{\tan }^{-1}}\left( \dfrac{4}{-2} \right)=\pi +{{\tan }^{-1}}\left( -2 \right).....\left( 1 \right)$.
Using equation (1), we can rewrite the given trigonometric expression as ${{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3={{\tan }^{-1}}2+\pi +{{\tan }^{-1}}\left( -2 \right).....\left( 2 \right)$.
We know the trigonometric identity ${{\tan }^{-1}}\left( -x \right) = -{{\tan }^{-1}}x$.
Substituting $x=2$ in the above equation, we have ${{\tan }^{-1}}\left( -2 \right)=-{{\tan }^{-1}}2.....\left( 3 \right)$.
Substituting equation (3) in equation (2), we have ${{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3={{\tan }^{-1}}2+\pi +{{\tan }^{-1}}\left( -2 \right)={{\tan }^{-1}}2+\pi -{{\tan }^{-1}}2$.
Simplifying the above expression, we have ${{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3={{\tan }^{-1}}2+\pi -{{\tan }^{-1}}2=\pi $.
Hence, the value of the trigonometric expression ${{\tan }^{-1}}1+{{\tan }^{-1}}2+{{\tan }^{-1}}3$ is $\pi $, which is option (c).
Trigonometric functions are real functions that relate any angle of a right-angled triangle to the ratios of any two of its sides. The widely used trigonometric functions are sine, cosine, and tangent. However, we can also use their reciprocals, i.e., cosecant, secant, and cotangent. We can use geometric definitions to express the value of these functions on various angles using unit circle (circle with radius 1). We also write these trigonometric functions as infinite series or as solutions to differential equations. Thus, allowing us to expand the domain of these functions from the real line to the complex plane.

Note: One must be careful while using the inverse trigonometric formula ${{\tan }^{-1}}x+{{\tan }^{-1}}y=\pi +{{\tan }^{-1}}\left( \dfrac{x+y}{1-xy} \right)$ if $xy>1$ as the formula shows different natures for different values of xy. We can also solve this question by calculating the value of each of the terms and then substituting it in the given expression.