Answer:
[tex]\mathrm{d. \csc \theta =\frac{5}{3}}[/tex]
Step-by-step explanation:
In a right triangle only, the tangent of an angle is equal to its opposite side divided by its adjacent side. If [tex]\cot\theta =\frac{4}{3}[/tex] as given in the problem, then [tex]\tan \theta =\frac{3}{4}[/tex], because [tex]\tan \theta =\frac{1}{\cot \theta}[/tex]. Therefore, the opposite side of angle theta is 3 and its adjacent side is 4. Thus, the hypotenuse of the triangle must be [tex]h=\sqrt{4^2+3^2}=\sqrt{25}=5[/tex].
The sine of an angle in a right triangle is equal to its opposite side divided by its hypotenuse.
Therefore, [tex]\sin \theta =\frac{3}{5}[/tex] (o/h) and since [tex]\sin \theta =\frac{1}{\csc \theta}[/tex], we have:
[tex]\csc \theta=\frac{1}{\sin \theta}=\frac{1}{\frac{3}{5}}=\boxed{\frac{5}{3}}[/tex]
