Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Inverse Trigonometric Functions

Question:

$tan^{-1}\left(\frac{c_1x - y}{c_1y+x}\right) +tan^{-1}\left(\frac{c_2-c_1}{1+c_2c_1}\right) +tan^{-1}\left(\frac{c_3-c_2}{1+c_23_2}\right)+ .......+ tan^{-1}\frac{1}{c_n}$ is equal to

Options:

$tan^{-1}\frac{y}{x}$

$tan^{-1}\frac{x}{y}$

$-tan^{-1}\frac{x}{y}$

none of these

Correct Answer:

$tan^{-1}\frac{x}{y}$

Explanation:

$tan^{-1}\left(\frac{c_1x - y}{c_1y+x}\right) +tan^{-1}\left(\frac{c_2-c_1}{1+c_2c_1}\right) +tan^{-1}\left(\frac{c_3-c_2}{1+c_3c_2}\right)+ .......+ tan^{-1}\frac{1}{c_n}$

$ = tan^{-1}\left(\frac{\frac{x}{y}-\frac{1}{c_1}}{1+\frac{x}{y}×\frac{1}{c_1}}\right) +tan^{-1}\left(\frac{\frac{1}{c_1}-\frac{1}{c_2}}{1+\frac{1}{c_1c_2}}\right)+tan^{-1}\left(\frac{\frac{1}{c_2}-\frac{1}{c_3}}{1+\frac{1}{c_2c_3}}\right) +.......+ tan^{-1}\frac{1}{c_n}$

$=\left(tan^{-1}\frac{x}{y}-tan^{-1}\frac{1}{c_1}\right) + \left(tan^{-1}\frac{1}{c_1}-tan^{-1}\frac{1}{c_2}\right) +\left(tan^{-1}\frac{1}{c_2}-tan^{-1}\frac{1}{c_3}\right) +.......+\left(tan^{-1}\frac{1}{c_{n-1}}-tan^{-1}\frac{1}{c_n}\right) + tan^{-1}\frac{1}{c_n} =tan^{-1}\frac{x}{y}$