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Algebra Questions

Question Number 10542 by FilupS last updated on 17/Feb/17

$$\mathrm{Prove}\:\mathrm{that}: \\$$$$\mathrm{tan}\left(\mathrm{sec}^{−\mathrm{1}} \left(\sqrt{\mathrm{tan}\left(\theta\right)}\right)\right)=\sqrt{\mathrm{tan}\left(\theta\right)}\sqrt{\mathrm{1}−\mathrm{cot}\left(\theta\right)} \\$$

Answered by mrW1 last updated on 17/Feb/17

$${let}\:\alpha=\mathrm{sec}^{−\mathrm{1}} \left(\sqrt{\mathrm{tan}\:\left(\theta\right)}\right) \\$$$$\Rightarrow\mathrm{sec}\:\alpha=\sqrt{\mathrm{tan}\:\left(\theta\right)} \\$$$$\Rightarrow\mathrm{sec}^{\mathrm{2}} \:\alpha=\mathrm{tan}\:\left(\theta\right) \\$$$$\mathrm{tan}\left(\mathrm{sec}^{−\mathrm{1}} \left(\sqrt{\mathrm{tan}\left(\theta\right)}\right)\right)=\mathrm{tan}\:\alpha=\sqrt{\mathrm{sec}^{\mathrm{2}} \:\alpha−\mathrm{1}}=\sqrt{\mathrm{t}{a}\mathrm{n}\:\left(\theta\right)−\mathrm{1}}=\sqrt{\mathrm{tan}\:\left(\theta\right)\left[\mathrm{1}−\frac{\mathrm{1}}{\mathrm{tan}\:\left(\theta\right)}\right]}=\sqrt{\mathrm{tan}\:\left(\theta\right)}\sqrt{\mathrm{1}−\mathrm{cot}\:\left(\theta\right)} \\$$$${proven}! \\$$