Question

Let theta be an angle in quadrant III such that cos theta=-3/5 . Find the exact values of csc theta and tan theta .

Answer 1

The angle in quadrant III that has a cosine of -3/5 is obtained by calculating the arccos of 3/5 and adding 180 to answer. This gives 233.13 degrees. The cosecant of this angle is -5/4 and the tangent is calculated to be 4/3. 

Answer 2

Answer:

$cosec\theta$=-$\frac{5}{4}$

$tan\theta$=$\frac{4}{3}$.

Step-by-step explanation:

Given, let $\theta$  be the angle in the III quadrant .

$cos\theta$=-$\frac{3}{5}$......    (given)

In III quadrant $cosec\theta$ is negative  and $tan\theta$ is positive .

because , $cosec\theta$=reciprocal of $sin\theta$

it means , $cosec\theta$=$\frac{1}{sin\theta}$

Therefore,  first we find value of $sin\theta$

 We know that

$sin^2\theta=1-cos^2\theta$

∴ $sin\theta=\sqrt{1-cos^2\theta}$

$sin\theta=\sqrt{1-(\frac{-3}{5} )^2$

$sin\theta=\sqrt{1-\frac{9}{25}}$

$sin\theta=\sqrt{\frac{16} {25}$

$sin\theta=\frac{-4}{5}$

Because $sin\theta$ lies in III quadrant and in III quadrant it is negative.

Now, we know that

$tan\theta=\sqrt{sec^2\theta-1}$

therefore,  first we find $sec\theta$

$sec\theta=\frac{1}{cos\theta}$

$sec\theta=-\frac{5}{3}$

$tan\theta=\sqrt{(\frac{5}{3})^2-1$

$tan\theta=\sqrt{\frac{25} {9}-1$

$tan\theta=\frac{4}{3}$

Because it lies in III quadrant, therefore it take positive.

Hence,$cosec\theta=-\frac{5}{4}$ and $tan\theta=\frac{4}{3}$