Question-173002

Question Number 173002 by Mikenice last updated on 04/Jul/22

Commented by kaivan.ahmadi last updated on 05/Jul/22

1 - {t g}^{2} θ = 1 - \frac{{s i n}^{2} θ}{{c o s}^{2} θ} = \frac{{c o s}^{2} θ - {s i n}^{2} θ}{{c o s}^{2} θ} =

\frac{1 - 2 {s i n}^{2} θ}{1 - {s i n}^{2} θ} = \frac{1 - 2 {(\frac{a - b}{a + b})}^{2}}{1 - {(\frac{a - b}{a + b})}^{2}} = \frac{{(a + b)}^{2} - 2 {(a - b)}^{2}}{{(a + b)}^{2} - {(a - b)}^{2}} =

\frac{- a^{2} + 6 a b - b^{2}}{4 a b}

Answered by CElcedricjunior last updated on 05/Jul/22

\sin θ = \frac{a - b}{a + b}

calculons 1 - \tan^{2} θ

1 - \tan^{2} θ = 1 - \frac{\sin^{2} θ}{1 - \sin^{2} θ}

= 1 - \frac{{(a - b)}^{2}}{{(a + b)}^{2}} \times \frac{{(a + b)}^{2}}{4 ab}

= \frac{- (a^{2} - 6 ab + b^{2})}{4 ab}

1 - \tan^{2} θ = \frac{- (a - 3 b - 2 b \sqrt{2}) (a - 3 b + 2 b \sqrt{2})}{4 ab}

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