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

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

$$1-{tg}^2\theta =1-\frac{{sin}^2\theta }{{cos}^2\theta }=\frac{{cos}^2\theta -{sin}^2\theta }{{cos}^2\theta }=$$$$\frac{1-2{sin}^2\theta }{1-{sin}^2\theta }=\frac{1-2{\left(\frac{a-b}{a+b}\right)}^2}{1-{\left(\frac{a-b}{a+b}\right)}^2}=\frac{{(a+b)}^2-2{(a-b)}^2}{{(a+b)}^2-{(a-b)}^2}=$$$$\frac{-a^2+6ab-b^2}{4ab}$$

Answered by CElcedricjunior last updated on 05/Jul/22

$$\mathbf{sin}\mathit{\theta }=\frac{\mathbf{a}-\mathbf{b}}{\mathbf{a}+\mathbf{b}}$$$$\mathbf{calculons}1-{\mathbf{tan}}^2\mathit{\theta }$$$$1-{\mathbf{tan}}^2\mathit{\theta }=1-\frac{{\mathbf{sin}}^2\mathit{\theta }}{1-{\mathbf{sin}}^2\mathit{\theta }}$$$$=1-\frac{{(\mathbf{a}-\mathbf{b})}^2}{{(\mathbf{a}+\mathbf{b})}^2}\times \frac{{(\mathbf{a}+\mathbf{b})}^2}{4\mathbf{ab}}$$$$=\frac{-({\mathbf{a}}^2-6\mathbf{ab}+{\mathbf{b}}^2)}{4\mathbf{ab}}$$$$1-{\mathbf{tan}}^2\mathit{\theta }=\frac{-(\mathbf{a}-3\mathbf{b}-2\mathbf{b}\sqrt{2})(\mathbf{a}-3\mathbf{b}+2\mathbf{b}\sqrt{2})}{4\mathbf{ab}}$$$$$$$$$$$$.........\mathcal{L}ec\acute{el}\stackrel{`}{ebre}cedricjunior............$$

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