Question

tan invese ( a+bx)divided by (b-ax),x ÷a

Answer 1

$\textbf{Consider,}$

$tan^{-1}[\frac{3a^2x-x^3}{a^3-3ax^2}]$

$\boxed{\begin{minipage}{3cm} \bf\text{Take, }x=a\;tan\theta\\\\\implies\theta=tan^{-1}\frac{x}{a} \end{minipage}}$

$=tan^{-1}[\frac{3a^2(a\;tan\theta)-a^3\;tan^3\theta}{a^3-3a(a^2\;tan^2\theta)}]$

$=tan^{-1}[\frac{3a^3\;tan\theta-a^3\;tan^3\theta}{a^3-3a^3\;tan^2\theta}]$

$=tan^{-1}[\frac{a^3(3\;tan\theta-tan^3\theta)}{a^3(1-3\;tan^2\theta)}]$

$=tan^{-1}[\frac{3\;tan\theta-tan^3\theta}{1-3\;tan^2\theta}]$

$\text{Using,}$

$\boxed{\bf\,tan\,3A=\frac{3\;tan\,A-tan^3A}{1-3\;tan^2\,A}}$

$=tan^{-1}[tan\;3\theta]$

$=3\theta$

$=3\;tan^{-1}\frac{x}{a}$