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  • find the value of <img alt="\tan \left[\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right]=" src="https://learn.careers360.com/latex-image/?%5Ctan%20%5Cleft%5B%5Ccos%20%5E%7B-1%7D%20%5Cfrac%7

find the value of \tan \left[\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right]=

Option: 1

{6 / 17}


Option: 2

{17 / 6}


Option: 3

{7 / 16}


Option: 4

{16 / 7}


Answers (1)

best_answer

 

 

Sum and difference of angles in terms of arctan (Part 2) -

Sum and difference of angles in terms of arctan (Part 2)

In the above formula replace y with -y, we will get

 

\\\mathrm{2.\;\;\tan^{-1}x-\tan^{-1}y=}\left\{\begin{matrix} \tan^{-1}\left ( \frac{x-y}{1+xy} \right), &\text{If xy}>-1 \\ \\ \pi+\tan^{-1}\left ( \frac{x-y}{1+xy} \right ), & \mathrm{If \;x>0,y<0\;and\;xy<-1}\\\\-\pi+\tan^{-1} \left ( \frac{x-y}{1+xy} \right ), & \mathrm{If \;x<0,y>0\;and\;xy<-1} \end{matrix}\right.


 

In the result 1, by replacing y with x , we will get

\\\mathrm{3.\;\;2\tan^{-1}x=}\left\{\begin{matrix} \tan^{-1}\left ( \frac{2x}{1-x^2} \right), &-1<x<1 \\ \\ \pi+\tan^{-1}\left ( \frac{2x}{1-x^2} \right ), & x>1\\\\-\pi+\tan^{-1} \left ( \frac{2x}{1-x^2} \right ), &x<-1 \end{matrix}\right.

-

 

 

 

\tan [\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}]=\tan [\tan ^{-1} \frac{\sqrt{(1-\frac{16}{25})}}{\frac{4}{5}}+\tan ^{-1} \frac{2}{3} ] \\ =\tan [\tan ^{-1}(\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4} \cdot \frac{2}{3}}] \\=\tan \cdot \tan ^{-1} \frac{17}{6}\\ =\frac{17}{6}

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