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Tan(sin^-1 (5/7)) solve in radians

User Bniwredyc
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1 Answer

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12 votes
Trigonometry


\begin{gathered} \tan (\sin ^(-1)((5)/(7)))=(\sin (\sin ^(-1)((5)/(7))))/(\cos (\sin ^(-1)((5)/(7)))) \\ =\frac{\sin (\sin ^(-1)((5)/(7)))}{\sqrt[]{1-\sin ^2(\sin ^(-1)((5)/(7)))}} \\ =\frac{(5)/(7)}{\sqrt[]{1-(\sin (\sin ^(-1)((5)/(7))))^2}} \\ =\frac{(5)/(7)}{\sqrt[]{1-((5)/(7))^2}} \end{gathered}

We know


\begin{gathered} \sqrt[]{1-((5)/(7))^2}=\sqrt[]{1-(25)/(49)} \\ =\sqrt[]{(49-25)/(49)} \\ =\sqrt[]{(24)/(49)} \\ =\frac{\sqrt[]{24}}{7} \\ =\frac{2\sqrt[]{6}}{7} \end{gathered}

Replacing in the previous equation


\frac{(5)/(7)}{\frac{2\sqrt[]{6}}{7}}=\frac{5}{2\sqrt[]{6}}

Multiplying both sides of the fraction


\frac{5}{2\sqrt[]{6}}=\frac{5\sqrt[]{6}}{2\cdot6}=\frac{5\sqrt[]{6}}{12}

User Veiset
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