Question

Using the attached link below.

Answer 1

$\text{Use the Pythagorean theorem:}\\\\r-hypotenuse\\\\r^2=7^2+5^2\\\\r^2=49+25\\\\r^2=74\to r=\sqrt{74}$

$\sin=\dfrac{opposite}{hypotenuse}\\\\\cos=\dfrac{adjacent}{hypotenuse}\\\\\tan=\dfrac{opposite}{adjacent}\\\\\cot=\dfrac{adjacent}{opposite}\\\\\text{We have}\\\\for\ the\ angle\ y:\\\text{opposite = 7}\\\text{adjacent = 5}\\\text{hypotenuse = }\ \sqrt{74}\\\\for\ the\ angle\ x:\\\text{opposite = 5}\\\text{adjacent = 7}\\\text{hypotenuse = }\ \sqrt{74}$

$\csc x=\dfrac{1}{\sin x}\\\\\sec x=\dfrac{1}{\cos x}$

$A.\\\\\sin x=\dfrac{5}{\sqrt{74}}=\dfrac{5\sqrt{74}}{74}\\\\\csc y=\dfrac{1}{\frac{7}{\sqrt{74}}}=\dfrac{\sqrt{74}}{7}\\\\B.\\\\\tan x=\dfrac{5}{7}\\\\\cot y=\dfrc{5}{7}\\\\C.\\\\\cos x=\dfrac{7}{\sqrt{74}}=\dfrac{7\sqrt{74}}{7}\\\\\sec y=\dfrac{1}{\frac{5}{\sqrt{74}}}=\dfrac{\sqrt{74}}{5}$

$D.\\\sin\theta=\dfrac{2}{3}\\\\\sin^2\theta+\cos^2\theta=1\to\left(\dfrac{2}{3}\right)^2+\cos^2\theta=1\\\\\dfrac{4}{9}+\cos^2\theta=1\qquad\text{subtract}\ \dfrac{4}{9}\ \text{from both sides}\\\\\cos^2\theta=\dfrac{5}{9}\to\cos\theta=\sqrt{\dfrac{5}{9}}\to\cos\theta=\dfrac{\sqrt5}{3}\\\\\tan\theta=\dfrac{\sin\theta}{\cos\theta}\to\tan\theta=\dfrac{\frac{2}{3}}{\frac{\sqrt5}{3}}=\dfrac{2}{3}\cdot\dfrac{3}{\sqrt5}=\dfrac{2}{\sqrt5}=\dfrac{3\sqrt5}{5}$