Question

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

Answer 1

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}`