Evaluate the expression. tan ( cos − 1 12 13 ) \tan\left(\cos^{-1}\frac{12}{13}\right) tan ( cos − 1 13 12 )

this problem, we're asked to evaluate the expression, the tangent of the inverse cosine of 12 over 13. Now getting right into our steps here, step number 1, we're gonna use that inside function and use that interval in order to identify our quadrant. Now here, our inside function is the inverse cosine, and the interval of my inverse cosine when dealing with angles is between 0 and pi. So I know that my right triangle is either going to go in quadrant 1 or quadrant 2 based on that interval. Now here looking at the sign of my argument, I'm taking the inverse cosine of positive 12 over 13, a positive value, and my values for cosine can only be positive in quadrant 1 within this interval. So that's where my right triangle is going to go. Now step 1 is done. Moving on to step 2, to actually draw our right triangle and then use our argument to label everything that we need to. Remember, we're drawing our triangle in quadrant 1 like we identified in step 1. So here I'm going to draw my right triangle, and I'm going to go ahead and label that inside angle that's measured from against the x axis. So here is my angle theta, and now I wanna use that argument 12 over 13 to label those sides. This argument tells me that the cosine of my angle theta is equal to 12 over 13. So in order for that to be true based on SOHCAHTOA, that means that my adjacent side has to be 12. My hypotenuse has to be 13. So from here, step number 2 is done. We can move on to step number 3, which is to use the the Pythagorean theorem to find that missing third side. Now our missing side is a leg length here, so we're either solving for a or b, doesn't matter which one. Setting up my Pythagorean theorem, a squared plus b squared equals c squared. I'm gonna be solving for a here, so I'm gonna leave that as a variable. I'm gonna add in that side length 12 squared, and that's equal to 13 squared. Now from here, I can simplify this. A squared +144 is equal to 169. Now subtracting 144 from both sides, that cancels out, leaving me with a squared is equal to 20 5. Now, final step in getting this side length here is taking the square root, which gives me a value of 5. Now, labeling that on my triangle here, my missing side length of 5 makes sense that it is a positive value here because whenever we go up in this first quadrant, my y values are positive. So from here, we can move on to step number 4 and actually use our triangle to evaluate that outside function. Now here, my outside function is the tangent. So here, I'm gonna take the tangent of my angle. In order to do that based on SOHCAHTOA, I need to take my opposite side, divide it by my adjacent one. So my opposite side here is 5. My adjacent side is 12. So this gives me my final answer. The tangent of the inverse cosine of 12 over 13 is 5 over 12. Thanks Thanks for watching and let me know if you have questions.

11. Inverse Trigonometric Functions and Basic Trig Equations

Evaluate Composite Trig Functions

Precalculus

11. Inverse Trigonometric Functions and Basic Trig Equations

Evaluate Composite Trig Functions

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Multiple Choice

Evaluate the expression.
$\tan\left(\cos^{-1}\frac{12}{13}\right)$

$\frac{12}{13}$

$\frac{12}{5}$

$\frac{5}{13}$

$\frac{5}{12}$

Verified step by step guidance

Understand the problem: We need to evaluate the expression $ \tan\left(\cos^{-1}\left(\frac{12}{13}\right)\right) $. This involves finding the tangent of an angle whose cosine is $ \frac{12}{13} $.

Recall the identity for inverse cosine: If $ \theta = \cos^{-1}\left(\frac{12}{13}\right) $, then $ \cos(\theta) = \frac{12}{13} $. We need to find $ \tan(\theta) $.

Use the Pythagorean identity: $ \sin^2(\theta) + \cos^2(\theta) = 1 $. Substitute $ \cos(\theta) = \frac{12}{13} $ into the identity to find $ \sin(\theta) $.

Calculate $ \sin(\theta) $: $ \sin^2(\theta) = 1 - \left(\frac{12}{13}\right)^2 $. Solve for $ \sin(\theta) $ by taking the square root of the result.

Find $ \tan(\theta) $: Use the definition $ \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} $. Substitute the values of $ \sin(\theta) $ and $ \cos(\theta) $ to find $ \tan(\theta) $.