if tan^-1 (alpha + ibeta)= x+iy then x is equal to?

Open in App

Solution

$\tan^{- 1} (α + iβ) = x + iy \tan (x + iy) = (α + iβ) \frac{tanx + \tan (iβ)}{1 - tanx \times \tan (iβ)} = (α + iβ) tanx + \tan (iβ) = (α + iβ) - (α + iβ) tanx \times \tan (iβ) tanx (1 + (α + iβ) \tan (iβ)) = (α + iβ) - \tan (iβ) tanx = \frac{(α + iβ) - \tan (iβ)}{(1 + (α + iβ) \tan (iβ))} = \frac{\tan [\tan^{- 1} (α + iβ)] - \tan (iβ)}{(1 + \tan [\tan^{- 1} (α + iβ)] \tan (iβ))} = \tan [\tan^{- 1} (α + iβ) - (iβ)] On comparision we get, x = \tan^{- 1} (α + iβ) - (iβ)$

Suggest Corrections

Similar questions

α + i β = {tan}^{- 1} z

z = x + i y

α

z

If $t a n α = \frac{x}{x + 1} a n d t a n β = \frac{1}{2 x + 1}$ , then $α + β$ is equal to

I f α

{tan}^{- 1} x + {cot}^{- 1} x = 2 {tan}^{- 1} x, t h e n α^{2}

α + i β = t a n^{- 1} (z), z = x + i y

α

z

{cot}^{- 1} [\sqrt{cos α}] - {tan}^{- 1} [\sqrt{cos α}] = x

sin x

Join BYJU'S Learning Program