Question

Given that sec θ = –5∕4 and θ is in quadrant II, find cot θ.A) 4∕3B) –3∕5C) –3∕4D) –4∕3

Answer 1

To answer this question, we can draw the situation in the unit circle. We know that the secant ratio is equal to:

`\sec \theta=\frac{\text{hyp}}{\text{adj}}=(1)/(\cos \theta)`

In the second quadrant, the function cosine is negative, and so is the secant function, and we know this because the adjacent side of the triangle is in the second quadrant (the x-values are negative there). Then, we can graph this as follows:

We can use the Pythagorean Theorem to find the side, x, necessary to find the cotangent function. Then, we have:

`(-4)^2+x^2=5^2\Rightarrow x^2=5^2-(-4)^2=25-16=9\Rightarrow\sqrt[]{x^2}=\sqrt[]{9}`

Then x = 3. (Using the same notation for the unit circle, we can say that y = 3.)

Since the trigonometric ratio for the cotangent function is given by:

`\cot \theta=(adj)/(opp)`

That is the adjacent side over the opposite side. The adjacent side of angle Θ is x = -4, and the opposite side is y = 3. Therefore, the cotΘ is:

`\cot \theta=(-4)/(3)\Rightarrow\cot \theta=-(4)/(3)`

To check this result, we know that:

`\sin ^2(x)+\cos ^2(x)=1`

Since

`\sec \theta=(1)/(\cos\theta)\Rightarrow\cos \theta=(1)/(\sec\theta)\Rightarrow cos\theta=(1)/(-(5)/(4))\Rightarrow\cos \theta=-(4)/(5)`

Then, we have:

`\sin ^2(x)+(-(4)/(5))^2=1\Rightarrow\sin ^2(x)=1-(-(4)/(5))^2\Rightarrow\sin ^2(x)=(9)/(25)`

Now, we have:

`\sin (x)=\sqrt[]{(9)/(25)}=\frac{\sqrt[]{9}}{\sqrt[]{25}}=(3)/(5)`

We also have that tangent is:

`\tan \theta=(\sin\theta)/(cos\theta)\Rightarrow\cot \theta=(1)/(\tan\theta)=(1)/((\sin\theta)/(\cos\theta))=(\cos \theta)/(\sin \theta)`

And finally

`\cot \theta=(-(4)/(5))/((3)/(5))=(-4\cdot5)/(5\cdot3)\Rightarrow\cot \theta=-(4)/(3)`

In summary, using two different methods, we found the same result. Therefore, cotΘ = -4/3 (option D).