Inverse Trigonometric Functions Question 219
Question: $ {{\tan }^{-1}}\frac{c_1x-y}{c_1y+x}+{{\tan }^{-1}}\frac{c_2-c_1}{1+c_2c_1}+ $ $ {{\tan }^{-1}}\frac{c_3-c_2}{1+c_3c_2}+…+{{\tan }^{-1}}\frac{1}{c_{n}}= $
Options:
A) $ {{\tan }^{-1}}\frac{y}{x} $
B) $ {{\tan }^{-1}}yx $
C) $ {{\tan }^{-1}}\frac{x}{y} $
D) $ {{\tan }^{-1}}(x-y) $
Show Answer
Answer:
Correct Answer: C
Solution:
$ {{\tan }^{-1}}( \frac{c_1x-y}{c_1y+x} )+{{\tan }^{-1}}( \frac{c_2-c_1}{1+c_2c_1} )+{{\tan }^{-1}}( \frac{c_3-c_2}{1+c_3c_2} )+ $
$ …..+{{\tan }^{-1}}\frac{1}{c_{n}} $ = $ {{\tan }^{-1}}( \frac{\frac{x}{y}-\frac{1}{c_1}}{1+\frac{x}{y}.\frac{1}{c_1}} )+{{\tan }^{-1}}( \frac{\frac{1}{c_1}-\frac{1}{c_2}}{1+\frac{1}{c_1c_2}} ) $
$ +{{\tan }^{-1}}( \frac{\frac{1}{c_2}-\frac{1}{c_3}}{1+\frac{1}{c_2c_3}} )+…….+{{\tan }^{-1}}\frac{1}{c_{n}} $
$ ={{\tan }^{-1}}\frac{x}{y}-{{\tan }^{-1}}\frac{1}{c_1}+{{\tan }^{-1}}\frac{1}{c_1}-{{\tan }^{-1}}\frac{1}{c_2}+{{\tan }^{-1}}\frac{1}{c_2} $
$ -{{\tan }^{-1}}\frac{1}{c_3}+…+{{\tan }^{-1}}\frac{1}{{c_{n-1}}}-{{\tan }^{-1}}\frac{1}{c_{n}}+{{\tan }^{-1}}\frac{1}{c_{n}} $ = $ {{\tan }^{-1}}( \frac{x}{y} ) $ .
BETA
