evaluate the expression.\ncot \\left(\\sin ^{-1}\\left(\\frac{18}{30}\\right)\\right)\n\\( \\bigcirc…

evaluate the expression.\ncot \\left(\\sin ^{-1}\\left(\\frac{18}{30}\\right)\\right)\n\\( \\bigcirc \\mathrm{a} \\cdot \\frac{4}{3} \\)\n\\( \\bigcirc \\mathrm{b} \\cdot \\frac{3}{5} \\)\n\\( \\bigcirc \\mathrm{c} \\cdot \\frac{5}{3} \\)\n\\( \\bigcirc \\mathrm{d} \\cdot \\frac{3}{4} \\)

evaluate the expression.\ncot \\left(\\sin ^{-1}\\left(\\frac{18}{30}\\right)\\right)\n\\( \\bigcirc \\mathrm{a} \\cdot \\frac{4}{3} \\)\n\\( \\bigcirc \\mathrm{b} \\cdot \\frac{3}{5} \\)\n\\( \\bigcirc \\mathrm{c} \\cdot \\frac{5}{3} \\)\n\\( \\bigcirc \\mathrm{d} \\cdot \\frac{3}{4} \\)

Answer

Explanation:

Step1: Let (\theta=\sin^{-1}(\frac{18}{30}))

By the definition of inverse - sine function, (\sin\theta=\frac{18}{30}=\frac{3}{5}), and (\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]).

Step2: Use the Pythagorean identity (\sin^{2}\theta+\cos^{2}\theta = 1)

We can find (\cos\theta). Substitute (\sin\theta=\frac{3}{5}) into the identity: (\cos^{2}\theta=1 - \sin^{2}\theta). Then (\cos^{2}\theta=1-\left(\frac{3}{5}\right)^{2}=1-\frac{9}{25}=\frac{16}{25}). Since (\theta\in[-\frac{\pi}{2},\frac{\pi}{2}]) and (\sin\theta=\frac{3}{5}>0), (\theta\in(0,\frac{\pi}{2})), so (\cos\theta=\frac{4}{5}).

Step3: Use the cotangent formula (\cot\theta=\frac{\cos\theta}{\sin\theta})

Substitute (\sin\theta = \frac{3}{5}) and (\cos\theta=\frac{4}{5}) into the formula. (\cot\theta=\frac{\frac{4}{5}}{\frac{3}{5}}).

Step4: Simplify the expression

(\cot\theta=\frac{4}{3}) (by canceling out the common factor of (5) in the numerator and denominator).

Answer:

A. (\frac{4}{3})