Question

tan⁻¹ 1/(α –1) = tan⁻¹(1/x) + tan⁻¹ 1/(α² – x + 1) को हल कीजिए।

Answer 1

$\Leftrightarrow x^2-( \alpha^2+1) x+\alpha^3-\alpha^2+\alpha=0$

Step-by-step explanation:

$tan^{-1}\frac{1}{\alpha-1} =tan^{-1}\frac{1}{x} +tan^{-1}\frac{1}{\alpha^2-x+1}$

$\Leftrightarrow tan^{-1}\frac{1}{\alpha-1} -tan^{-1}\frac{1}{x} =tan^{-1}\frac{1}{\alpha^2-x+1}$

$\Leftrightarrow tan^{-1}\frac{\frac{1}{\alpha -1}-\frac{1}{x} }{1+\frac{1}{\alpha -1}.\frac{1}{x} } =tan^{-1}\frac{1}{\alpha^2-x+1}$

$\Leftrightarrow tan^{-1}\frac{x-\alpha+1}{\alpha x-x+1} =tan^{-1}\frac{1}{\alpha^2-x+1}$

$\Leftrightarrow \frac{x-\alpha+1}{\alpha x-x+1} =\frac{1}{\alpha^2-x+1}$

$\Leftrightarrow( {x-\alpha+1}){(\alpha^2-x+1}) ={1}}{(\alpha x-x+1)}$

$\Leftrightarrow \alpha^2 x-\alpha^3+\alpha^2-x^2+\alpha x-x+x-\alpha+1=\alpha x-x+1$

$\Leftrightarrow \alpha^2 x-\alpha^3+\alpha^2-x^2-\alpha+x=0$

$\Leftrightarrow( \alpha^2+1) x-\alpha^3+\alpha^2-x^2-\alpha=0$

$\Leftrightarrow x^2-( \alpha^2+1) x+\alpha^3-\alpha^2+\alpha=0$