Find $\tan t$ given that $\sin t = \frac { 3 } { 5 }$ and $\mathrm { cot } t

Question

Find $\tan t$ given that $\sin t = \frac { 3 } { 5 }$ and $\mathrm { cot } t < 0$
A) $\tan t = - \frac { 3 } { 4 }$

B) $\tan t = - \frac { 3 } { 5 }$

C) $\tan t = - 1$

D) $\tan t = \frac { 3 } { 4 }$

E) $\tan t = - \frac { \sqrt { 3 } } { 2 }$

Quizplus · Accepted Answer

Given $\sin t = \frac{3}{5}$, we can find $\cos t$ using the Pythagorean identity $\sin^2 t + \cos^2 t = 1$. Solving for $\cos t$, we get $\cos t = \pm\frac{4}{5}$. Since $\cot t = \frac{1}{\tan t} < 0$, $\tan t$ and $\cot t$ must have opposite signs, implying $\cos t < 0$ (because $\sin t > 0$). Thus, $\cos t = -\frac{4}{5}$, and $\tan t = \frac{\sin t}{\cos t} = \frac{3/5}{-4/5} = -\frac{3}{4}$.

Find tan⁡t\tan ttant Given That sin⁡t=35\sin t = \frac { 3 } { 5 }sint=53​

Find $\tan t$ Given That $\sin t = \frac { 3 } { 5 }$