2. given \\( \\sin \\theta = \\frac { 3 } { 10 } \\), where \\( 90 ^ { \\circ } \\leq \\theta \\leq 180 ^ {…

Explanation:

Step1: Find (\cos\theta)

Use the identity (\sin^{2}\theta+\cos^{2}\theta = 1). Given (\sin\theta=\frac{3}{10}), then (\cos^{2}\theta=1 - \sin^{2}\theta=1-\left(\frac{3}{10}\right)^{2}=1-\frac{9}{100}=\frac{91}{100}). Since (90^{\circ}\leq\theta\leq180^{\circ}), (\cos\theta=-\frac{\sqrt{91}}{10}).

Step2: Find (\tan\theta)

Use the identity (\tan\theta=\frac{\sin\theta}{\cos\theta}). (\tan\theta=\frac{\frac{3}{10}}{-\frac{\sqrt{91}}{10}}=-\frac{3}{\sqrt{91}}=-\frac{3\sqrt{91}}{91}).

Step3: Find (\csc\theta)

Use the identity (\csc\theta=\frac{1}{\sin\theta}). (\csc\theta=\frac{1}{\frac{3}{10}}=\frac{10}{3}).

Step4: Find (\sec\theta)

Use the identity (\sec\theta=\frac{1}{\cos\theta}). (\sec\theta=\frac{1}{-\frac{\sqrt{91}}{10}}=-\frac{10}{\sqrt{91}}=-\frac{10\sqrt{91}}{91}).

Step5: Find (\cot\theta)

Use the identity (\cot\theta=\frac{\cos\theta}{\sin\theta}). (\cot\theta=\frac{-\frac{\sqrt{91}}{10}}{\frac{3}{10}}=-\frac{\sqrt{91}}{3}).

Step6: Find (\theta) (for part b)

Use (\theta=\sin^{- 1}\left(\frac{3}{10}\right)) in the second - quadrant. (\theta = 180^{\circ}-\sin^{-1}\left(\frac{3}{10}\right)). (\sin^{-1}\left(\frac{3}{10}\right)\approx17.46^{\circ}), so (\theta\approx180 - 17.46=163^{\circ}).

Answer:

a) (\cos\theta=-\frac{\sqrt{91}}{10}), (\tan\theta=-\frac{3\sqrt{91}}{91}), (\csc\theta=\frac{10}{3}), (\sec\theta=-\frac{10\sqrt{91}}{91}), (\cot\theta=-\frac{\sqrt{91}}{3}) b) (\theta\approx163^{\circ})