The solution of sin^(−1) (((2a)/(1+a^2 )))−cos^(−1) (((1−b^2 )/(1+b^2 )))=tan^(−1) (((2x)/(1−x^2 ))) is


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Question Number 43179 by gunawan last updated on 08/Sep/18

$The solution of$ $\sin^{- 1} (\frac{2 a}{1 + a^{2}}) - \cos^{- 1} (\frac{1 - b^{2}}{1 + b^{2}}) = \tan^{- 1} (\frac{2 x}{1 - x^{2}}) is$
	Answered by tanmay.chaudhury50@gmail.com last updated on 08/Sep/18

	$a = t a n α b = t a n β x = t a n γ$ ${s i n}^{- 1} (\frac{2 t a n α}{1 + {t a n}^{2} α}) - {c o s}^{- 1} (\frac{1 - {t a n}^{2} β}{1 + {t a n}^{2} β}) = {t a n}^{- 1} (\frac{2 t a n γ}{1 - {t a n}^{2} γ})$ ${s i n}^{- 1} (s i n 2 α) - {c o s}^{- 1} (c o s 2 β) = {t a n}^{- 1} (t a n 2 γ)$ $2 α - 2 β = 2 γ$ $α - β = γ$ $t a n (α - β) = t a n γ$ $\frac{t a n α - t a n β}{1 + t a n α t a n β} = t a n γ$ $\frac{a - b}{1 + a b} = x$
	Commented by gunawan last updated on 08/Sep/18

	$thank you very much Sir$
	Commented by tanmay.chaudhury50@gmail.com last updated on 08/Sep/18

	$y o u a r e w e l c o m e . . . e v e r y b o d y i s u p l o a d i n g s e l f$ $p h o t o . . . s o y o u p l s . . .$
	Commented by gunawan last updated on 08/Sep/18

	Commented by tanmay.chaudhury50@gmail.com last updated on 08/Sep/18

	$e x c e l l e n t . . .$
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