Answer:
Second answer
Explanation:
We are given
and
. What we have to find are
and
.
First, convert
to
via trigonometric identity. That gives us a new equation in form of
:
![\displaystyle \large{(1)/(\cos \theta) = -3}](https://img.qammunity.org/2023/formulas/mathematics/high-school/gtx88mzhd93cum8k9dmo3lnrvjdw3djxty.png)
Multiply
both sides to get rid of the denominator.
![\displaystyle \large{(1)/(\cos \theta) \cdot \cos \theta = -3 \cos \theta}\\\displaystyle \large{1=-3 \cos \theta}](https://img.qammunity.org/2023/formulas/mathematics/high-school/8vi9fwlcqhhxvgkzz4cdky43bwiaio5yxp.png)
Then divide both sides by -3 to get
.
Hence,
__________________________________________________________
Next, to find
, convert it to
via trigonometric identity. Then we have to convert
to
via another trigonometric identity. That gives us:
![\displaystyle \large{(1)/((\sin \theta)/(\cos \theta))}\\\displaystyle \large{(\cos \theta)/(\sin \theta)](https://img.qammunity.org/2023/formulas/mathematics/high-school/1hs9ps6gksvhrd0r60bxwva1rr5sm882c3.png)
It seems that we do not know what
is but we can find it by using the identity
for
.
From
then
.
Therefore:
![\displaystyle \large{\sin \theta=\sqrt{1-(1)/(9)}}\\\displaystyle \large{\sin \theta = \sqrt{(9)/(9)-(1)/(9)}}\\\displaystyle \large{\sin \theta = \sqrt{(8)/(9)}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/u26b2ls295s37m0ng97xp3g564d6sncl4f.png)
Then use the surd property to evaluate the square root.
Hence,
![\displaystyle \large{\boxed{\sin \theta=(2√(2))/(3)}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/cngoz91yflq5tbmsbuby6o2bxh69zwjv0j.png)
Now that we know what
is. We can evaluate
which is another form or identity of
.
From the boxed values of
and
:-
![\displaystyle \large{\cot \theta = (\cos \theta)/(\sin \theta)}\\\displaystyle \large{\cot \theta = (-(1)/(3))/((2√(2))/(3))}\\\displaystyle \large{\cot \theta=-(1)/(3) \cdot (3)/(2√(2))}\\\displaystyle \large{\cot \theta=-(1)/(2√(2))](https://img.qammunity.org/2023/formulas/mathematics/high-school/p4vt2bzc7ewcsrc9wtb562fvp7p3ggp26t.png)
Then rationalize the value by multiplying both numerator and denominator with the denominator.
![\displaystyle \large{\cot \theta = -(1 \cdot 2√(2))/(2√(2) \cdot 2√(2))}\\\displaystyle \large{\cot \theta = -(2√(2))/(8)}\\\displaystyle \large{\cot \theta = -(√(2))/(4)}](https://img.qammunity.org/2023/formulas/mathematics/high-school/wp6btlku7qb069w3lcamlpa8u7tvstj5bl.png)
Hence,
![\displaystyle \large{\boxed{\cot \theta = -(√(2))/(4)}}](https://img.qammunity.org/2023/formulas/mathematics/high-school/84mvd5iubl2mh6of4h3xhgy0becrjrmfr7.png)
Therefore, the second choice is the answer.
__________________________________________________________
Summary
![\displaystyle \large{\sec \theta = (1)/(\cos \theta)}\\ \displaystyle \large{\cot \theta = (1)/(\tan \theta) = (\cos \theta)/(\sin \theta)}\\ \displaystyle \large{\sin \theta = √(1-\cos ^2 \theta) \ \ \ (\sin \theta > 0)}\\ \displaystyle \large{\tan \theta = (\sin \theta)/(\cos \theta)}](https://img.qammunity.org/2023/formulas/mathematics/high-school/v3s1pbjrzyww4gz1qhxgnvcol6ixuylcgp.png)
![\displaystyle \large{\sqrt{(x)/(y)} = (√(x))/(√(y))}](https://img.qammunity.org/2023/formulas/mathematics/high-school/dbq3okwt5hzgrwv1htmwc7hfjepw3ogyrw.png)
Let me know in the comment if you have any questions regarding this question or for clarification! Hope this helps as well.