If $\operatorname{cosec} \theta=\frac{5}{3}$ and $\theta$ lies in Quadrant II, find the values of all the other five trigonometric functions.
Given: $\operatorname{cosec} \theta=\frac{5}{3}$
Since, θ is in IInd Quadrant. So, cos and tan will be negative but sin will be positive.
Now, we know that
$\sin \theta=\frac{1}{\operatorname{cosec} \theta}$
Putting the values, we get
$\sin \theta=\frac{1}{\frac{5}{3}}$
$\sin \theta=\frac{3}{5}$ …(i)
We know that,
$\sin ^{2} \theta+\cos ^{2} \theta=1$
Putting the values, we get
$\left(\frac{3}{5}\right)^{2}+\cos ^{2} \theta=1$ [from (i)]
$\Rightarrow \frac{9}{25}+\cos ^{2} \theta=1$
$\Rightarrow \cos ^{2} \theta=1-\frac{9}{25}$
$\Rightarrow \cos ^{2} \theta=\frac{25-9}{25}$
$\Rightarrow \cos ^{2} \theta=\frac{16}{25}$
$\Rightarrow \cos \theta=\sqrt{\frac{16}{25}}$
$\Rightarrow \cos \theta=\pm \frac{4}{5}$
Since, $\theta$ in IInd quadrant and $\cos \theta$ is negative in II $^{\text {nd }}$ quadrant
$\therefore \cos \theta=-\frac{4}{5}$
Now,
$\tan \theta=\frac{\sin \theta}{\cos \theta}$
Putting the values, we get
$\tan \theta=\frac{\frac{3}{5}}{-\frac{4}{5}}$
$=\frac{3}{5} \times\left(-\frac{5}{4}\right)$
$=-\frac{3}{4}$
Now,
$\sec \theta=\frac{1}{\cos \theta}$
Putting the values, we get
$\sec \theta=\frac{1}{-\frac{4}{5}}$
$=-\frac{5}{4}$
Now,
$\cot \theta=\frac{1}{\tan \theta}$
Putting the values, we get
$\cot \theta=\frac{1}{-\frac{3}{4}}$
$=-\frac{4}{3}$
Hence, the values of other trigonometric Functions are
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