The-solution-of-sin-1-2a-1-a-2-cos-1-1-b-2-1-b-2-tan-1-2x-1-x-2-is-

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 \dots 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 \dots s o y o u p l s \dots

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 \dots