Respuesta :

Trigonometry

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

We know

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

Replacing in the previous equation

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

Multiplying both sides of the fraction

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

Otras preguntas