Question

Evaluate ${\tan }^{-1}\left(\frac{{\mathrm{c}}_1\mathrm{x}-\mathrm{y}}{{\mathrm{c}}_1\mathrm{y}+\mathrm{x}}\right)+{\tan }^{-1}\left(\frac{{\mathrm{c}}_2-{\mathrm{c}}_1}{1+{\mathrm{c}}_2{\mathrm{c}}_1}\right)+{\tan }^{-1}\left(\frac{{\mathrm{c}}_3-{\mathrm{c}}_2}{1+{\mathrm{c}}_2{\mathrm{c}}_3}\right)+\dots +{\tan }^{-1}\left(\frac{1}{{\mathrm{c}}_{\mathrm{n}}}\right):$

• A. ${\tan }^{-1}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)$
• B. ${\tan }^{-1}\mathrm{yx}$
• C. ${\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)$
• D. ${\tan }^{-1}\left(\mathrm{x}-\mathrm{y}\right)$

Answer 1

The correct option is C

${\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)$

Explanation for correct option

Simplifying the given expression:

Given expression is: ${\tan }^{-1}\left(\frac{{\mathrm{c}}_1\mathrm{x}-\mathrm{y}}{{\mathrm{c}}_1\mathrm{y}+\mathrm{x}}\right)+{\tan }^{-1}\left(\frac{{\mathrm{c}}_2-{\mathrm{c}}_1}{1+{\mathrm{c}}_2{\mathrm{c}}_1}\right)+{\tan }^{-1}\left(\frac{{\mathrm{c}}_3-{\mathrm{c}}_2}{1+{\mathrm{c}}_2{\mathrm{c}}_3}\right)+\dots +{\tan }^{-1}\left(\frac{1}{{\mathrm{c}}_{\mathrm{n}}}\right)$

Firstly, will use the first term to equate,

${\tan }^{-1}\left(\frac{{\mathrm{c}}_1\mathrm{x}-\mathrm{y}}{{\mathrm{c}}_1\mathrm{y}+\mathrm{x}}\right)={\tan }^{-1}\left(\frac{{\displaystyle \frac{{\mathrm{c}}_1\mathrm{x}}{{\mathrm{c}}_1\mathrm{y}}}-{\displaystyle \frac{\mathrm{y}}{{\mathrm{c}}_1\mathrm{y}}}}{{\displaystyle \frac{{\mathrm{c}}_1\mathrm{y}}{{\mathrm{c}}_1\mathrm{y}}}+{\displaystyle \frac{\mathrm{x}}{{\mathrm{c}}_1\mathrm{y}}}}\right)$

$={\tan }^{-1}\left(\frac{{\displaystyle \frac{\mathrm{x}}{\mathrm{y}}-\frac{1}{{\mathrm{c}}_1}}}{1+{\displaystyle \frac{\mathrm{x}}{{\mathrm{c}}_1\mathrm{y}}}}\right).......\left(\mathrm{i}\right)$

We know that:

${\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{a}\mathbf{-}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\mathbf{b}\mathbf{=}{\mathbf{tan}}^{\mathbf{-}\mathbf{1}}\left[\frac{a-b}{1+\mathrm{ab}}\right]$

Applying the above formula in equation $\left(\mathrm{i}\right)$ we get::

$\therefore {\tan }^{-1}\left[\frac{\frac{\mathrm{x}}{\mathrm{y}}-\frac{1}{{\mathrm{c}}_1}}{1+\frac{\mathrm{x}}{\mathrm{y}}·\frac{1}{{\mathrm{c}}_1}}\right]={\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)-{\tan }^{-1}\left(\frac{1}{{\mathrm{c}}_1}\right)$

Now,

${\tan }^{-1}\left(\frac{{\mathrm{c}}_2-{\mathrm{c}}_1}{1+{\mathrm{c}}_2{\mathrm{c}}_1}\right)={\tan }^{-1}\left[\frac{\frac{1}{{\mathrm{c}}_1}-\frac{1}{{\mathrm{c}}_2}}{1+\frac{1}{{\mathrm{c}}_1{\mathrm{c}}_2}}\right]$

$={\tan }^{-1}\left(\frac{1}{{\mathrm{c}}_1}\right)-{\tan }^{-1}\left(\frac{1}{{\mathrm{c}}_2}\right)$

Continuing further, the expression becomes:

${\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)-{\tan }^{-1}\left(\frac{1}{{\mathrm{c}}_1}\right)+{\tan }^{-1}\left(\frac{1}{{\mathrm{c}}_1}\right)-{\tan }^{-1}\left(\frac{1}{{\mathrm{c}}_2}\right)+{\tan }^{-1}\left(\frac{1}{{\mathrm{c}}_2}\right)-{\tan }^{-1}\left(\frac{1}{{\mathrm{c}}_3}\right)+.......+-{\tan }^{-1}\left(\frac{1}{{\mathrm{c}}_{\mathrm{n}}}\right)={\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)$

Therefore, the expression reduces to ${\tan }^{-1}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)$.

Hence, the correct answer is option (C).