Answer:
[tex]\sin(\theta)=-12/13\text{ and } \csc(\theta)=-13/12\\\cos(\theta)=-5/13\text{ and } \sec(\theta)=-13/5\\\tan(\theta)=12/5\text{ and } \cot(\theta)=5/12[/tex]
Step-by-step explanation:
So we know that:
[tex]\sin(\theta)=-12/13\text{ and } \sec(\theta)<0[/tex]
This tells us that our angle θ is in Quadrant III. This is because, recall ASTC. In QI, everything is positive. This is not the case here since sine and secant are both negative.
In QII, only sine is positive. Since sine is negative, θ can't be in QII.
In QIII, only tangent (and cotangent) is positive. All other are negative. So, θ is in QIII.
And just to check, in QIV, only cosine (and secant) is positive. Since we are told that secant is less than 0 (in other words, negative), θ can't be in QIV.
Now that we know that θ is in QIII, we know that sine and cosecant is negative; cosine and secant is negative; and tangent and cotangent is positive.
Now, let's find the ratios. Remember what sine tells us. Sine gives us the ratio of the opposite side to the hypotenuse. With this information, let's find the adjacent side:
So, the opposite is 12 (ignore the negative for now) and the hypotenuse is 13. Therefore, we can use the Pythagorean Theorem:
[tex]a^2+b^2=c^2[/tex]
Substitute 12 for a and 13 for c:
[tex]12^2+b^2=c^2[/tex]
Now, solve for b to find the adjacent side. Square the values:
[tex]144+b^2=169[/tex]
Subtract 144 from both sides:
[tex]b^2=25[/tex]
Square root:
[tex]b=\pm 5[/tex]
We can ignore the negative and just take the positive. While the answer is technically -5, we don't need to do that. So, our adjacent side is 5.
Therefore, with respect to angle θ, our opposite is 12, our adjacent is 5, and our hypotenuse is 13. With this, we can compute the other trig ratios. Let's start with the 3 main ones:
Sine:
[tex]\sin(\theta)[/tex]
As given, this is -12/13. So:
[tex]\sin(\theta)=-12/13[/tex]
Cosine:
[tex]\cos(\theta)=adj/hyp[/tex]
Substitute 5 for the adjacent and 13 for the hypotenuse. So:
[tex]\cos(\theta)=5/13[/tex]
And since our angle is in QIII, we add a negative:
[tex]\cos(\theta)=-5/13[/tex]
Tangent:
[tex]\tan(\theta)=opp/adj[/tex]
Substitute 12 for opposite and 5 for adjacent. So:
[tex]\tan(\theta)=12/5[/tex]
And since our angle is in QIII, tangent stays positive.
To find the other three, simply flip the previous fractions:
[tex]\csc(\theta)=-13/12\\\sec(\theta)=-13/5\\\cot(\theta)=5/12[/tex]
