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Question Number 202976 by ajfour last updated on 06/Jan/24

Commented by ajfour last updated on 06/Jan/24

$$If\tan \alpha =t,findh.$$

Answered by mr W last updated on 06/Jan/24

Commented by mr W last updated on 06/Jan/24

$$b^2=(1-h)(1+h)=1-h^2$$$$\Rightarrow b=\sqrt{1-h^2}$$$$\tan {\alpha }_1=\frac{h}{b}$$$$\tan {\alpha }_2=\frac{1-h}{b}$$$$\alpha ={\alpha }_1+{\alpha }_2$$$$\tan \alpha =\frac{\frac{h}{b}+\frac{1-h}{b}}{1-\frac{h(1-h)}{b^2}}=\sqrt{\frac{1+h}{1-h}}=t$$$$\Rightarrow h=\frac{t^2-1}{t^2+1}$$

Commented by Frix last updated on 06/Jan/24

$$\mathrm{Yes}.\mathrm{And}\mathrm{with}t=\tan \alpha \mathrm{we}\mathrm{have}$$$$h=-\cos 2\alpha =1-{2\cos }^2\alpha$$

Answered by a.lgnaoui last updated on 06/Jan/24

$$2\mathit{\alpha }+\lambda =\pi \lambda =\pi -2\mathit{\alpha }$$$${\mathbf{AB}}^2=2(1+\cos 2\mathit{\alpha }){\mathbf{OA}}^2=1$$$${\mathbf{AB}}^2-{\mathbf{OA}}^2={(1-\mathbf{h})}^2-{\mathbf{h}}^2$$$$2(1+\cos 2\mathit{\alpha })-1=1-2\mathbf{h}$$$$2\cos 2\mathit{\alpha }=-2\mathrm{h}$$$$\cos 2\mathit{\alpha }=-\mathbf{h}$$$$\Rightarrow \mathbf{h}=\frac{{\mathbf{t}}^2-1}{1+{\mathbf{t}}^2}$$

Commented by a.lgnaoui last updated on 06/Jan/24

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