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In a right triangle, 0 is an acute and sin 0=2/7. Evaluate the other five trigonometric functions of 0.

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gmany

[tex] \sin\theta=\dfrac{2}{7}\\\\\text{use}\ \sin^2\theta+\cos^2\theta=1\to\left(\dfrac{2}{7}\right)^2+\cos^2\theta=1\\\\\dfrac{4}{49}+\cos^2\theta=1\ \ \ |-\dfrac{4}{49}\\\\\cos^2\theta=\dfrac{45}{49}\to\cos\theta=\sqrt{\dfrac{45}{49}}\\\\\cos\theta=\dfrac{\sqrt{45}}{\sqrt{49}}\\\\\cos\theta=\dfrac{\sqrt{9\cdot5}}{7}\\\\\cos\theta=\dfrac{3\sqrt5}{7} [/tex]

[tex] \text{use}\ \tan\theta=\dfrac{\sin\theta}{\cos\theta}\to\tan\theta=\dfrac{\frac{2}{7}}{\frac{3\sqrt5}{7}}=\dfrac{2}{7}\cdot\dfrac{7}{3\sqrt5}=\dfrac{2}{3\sqrt5}\cdot\dfrac{\sqrt5}{\sqrt5}\\\\=\dfrac{2\sqrt5}{3\cdot5}=\dfrac{2\sqrt5}{15} [/tex]

[tex] \text{use}\ \cot\theta=\dfrac{\cos\theta}{\sin\theta}\to\cot\theta=\dfrac{\frac{3\sqrt5}{7}}{\frac{2}{7}}=\dfrac{3\sqrt5}{7}\cdot\dfrac{7}{2}=\dfrac{3\sqrt5}{2} [/tex]

[tex] \text{use}\ \sec\theta=\dfrac{1}{\cos\theta}\to\sec\theta=\dfrac{1}{\frac{3\sqrt5}{7}}=\dfrac{7}{3\sqrt5}\cdot\dfrac{\sqrt5}{\sqrt5}=\dfrac{7\sqrt5}{3\cdot5}=\dfrac{7\sqrt5}{15} [/tex]

[tex] \text{use}\ \csc\theta=\dfrac{1}{\sin\theta}\to\csc\theta=\dfrac{1}{\frac{2}{7}}=\dfrac{7}{2} [/tex]