Question

y=tan^-1(x/1+6x^2)+cot^-1(1-10x^2/7x)​

Answer 1

$\textbf{To simplify:}$

$y=tan^{-1}[\frac{x}{1+6x^2}]+cot^{-1}[\frac{1-10x^2}{7x}]$

$\textbf{Solution:}$

$\text{Consider,}$

$y=tan^{-1}[\frac{x}{1+6x^2}]+cot^{-1}[\frac{1-10x^2}{7x}]$

$\text{Using,}$

$\boxed{\bf\,cot^{-1}x=tan^{-1}(\frac{1}{x})}$

$y=tan^{-1}[\frac{x}{1+6x^2}]+tan^{-1}[\frac{7x}{1-10x^2}]$

$\text{Using,}$

$\boxed{\bf\,tan^{-1}x+tan^{-1}y=tan^{-1}(\frac{x+y}{1-xy})}$

$y=tan^{-1}[\dfrac{\frac{x}{1+6x^2}+\frac{7x}{1-10x^2}}{1-\frac{x}{1+6x^2}\,\frac{7x}{1-10x^2}}]$

$y=tan^{-1}[\dfrac{\frac{x(1-10x^2)+7x(1+6x^2)}{(1+6x^2)(1-10x^2)}}{1-\frac{7x^2}{(1+6x^2)(1-10x^2)}}]$

$y=tan^{-1}[\dfrac{\frac{x-10x^3+7x+42x^3}{(1+6x^2)(1-10x^2)}}{\frac{(1+6x^2)(1-10x^2)-7x^2}{(1+6x^2)(1-10x^2)}}]$

$y=tan^{-1}[\dfrac{x-10x^3+7x+42x^3}{1-10x^2+6x^2-60x^4-7x^2}]$

$y=tan^{-1}[\dfrac{32x^3+8x}{-60x^4-11x^2+1}]$

$y=tan^{-1}[\dfrac{8x(4x^2+1)}{-(60x^4+11x^2-1)}]$

$y=tan^{-1}[\dfrac{8x(4x^2+1)}{-(60x^4+15x^2-4x^2-1)}]$

$y=tan^{-1}[\dfrac{8x(4x^2+1)}{-(15x^2(4x^2+1)-1(4x^2+1)}]$

$y=tan^{-1}[\dfrac{8x(4x^2+1)}{-(15x^2-1)(4x^2+1)}]$

$y=tan^{-1}[\dfrac{8x}{-(15x^2-1)}]$

$\implies\boxed{\bf\,y=tan^{-1}[\dfrac{8x}{1-15x^2)}]}$