Expand the expression using the sum & difference identities and simplify. tan ( − θ − π 2 ) \tan\left(-\theta-\frac{\pi}{2}\right) tan ( − θ − 2 π )

In this problem, we're asked to expand the expression using our sum and difference identities and then simplify. Now the expression that we're given here is the tangent of negative theta minus pi over 2. Now your first instinct here might be to expand this using your difference formula for tangent. So let's go ahead and see what happens if we do that. Now expanding this out, I end up with the tangent of negative theta minus the tangent of pi over 2 over 1 minuteus or 1 plus the tangent of negative theta, again, times the tangent of pi over 2. But what is the tangent of pi over 2? Well, the tangent of pi over 2 is an undefined value, so we're not going to get anywhere with this. So instead we're going to rewrite this in terms of sine and cosine. Then we can use our sine and cosine, sum and difference formulas in order to get to an answer. So let's go ahead and restart and do it using that method. So remember that whenever we end up with an undefined value for tangent, if either the tangent of a or the tangent of b is undefined, we want to go ahead and rewrite this in terms of sine and cosine. So the tangent of negative theta minus pi over 2 is equal to the sine of negative theta minus pi over 2 over the cosine of negative theta minus pi over 2. Now we can use our difference formulas for sine and cosine here. So let's go ahead and expand this out. Now in my numerator, I'll end up with the sine of negative theta times the cosine of pi over 2 minuteus the cosine of negative theta times the sine of pi over 2. So that's my numerator fully expanded. Then in my denominator, we're using our difference formula for cosine in order to get the cosine of negative theta times the cosine of pi over 2, plus the sine of negative theta, times the sine of pi over 2. Now all of these trig values, cosine of pi over 2 and sine of pi over 2, we know these values. Now the cosine of pi over 2 is equal to 0. So this term is 0 making this entire term go away. Now, that will happen in my denominator as well because the cosine of pi over 2 is still 0 there. Now the sine of pi over 2 is equal to 1, so that will just leave me with these two terms here. So here I'm left in my numerator with negative cosine of negative theta and then my denominator of sine of negative theta, but we can simplify this further because we know our even odd identities. The cosine of negative theta is simply the cosine of theta. So in my numerator I'm left with negative cosine of theta. Then in my denominator the sine of negative theta is negative sine of theta So here I have a negative cosine divided by negative sine. So what does this give me as a final answer? Well, it gives me positive cotangent of theta. Now that's my final answer fully simplified. Thanks for watching and I'll see you in the next one.

Expand the expression using the sum & difference identities and simplify. tan ⁡ ( − θ − π 2 ) \tan\left(-\theta-\frac{\pi}{2}\right) tan ( − θ − 2 π )

− tan ⁡ θ -\tan\theta − tan θ

− cot ⁡ θ -\cot\theta − cot θ

12. Trigonometric Identities

Sum and Difference Identities

Precalculus

12. Trigonometric Identities

Sum and Difference Identities

Struggling with Precalculus?

Join thousands of students who trust us to help them ace their exams!Watch the first video

Multiple Choice

Expand the expression using the sum & difference identities and simplify.
$\tan\left(-\theta-\frac{\pi}{2}\right)$

$-\tan\theta$

$\tan\theta$

$-\cot\theta$

$\cot\theta$

Verified step by step guidance

Recognize that the expression involves the tangent of a sum of angles. The angle is $-\theta - \frac{\pi}{2}$.

Use the tangent sum identity: $ \tan(a + b) = \frac{\tan a + \tan b}{1 - \tan a \tan b} $. Here, $ a = -\theta $ and $ b = -\frac{\pi}{2} $.

Substitute $ a = -\theta $ and $ b = -\frac{\pi}{2} $ into the identity: $ \tan(-\theta - \frac{\pi}{2}) = \frac{\tan(-\theta) + \tan(-\frac{\pi}{2})}{1 - \tan(-\theta) \tan(-\frac{\pi}{2})} $.

Recall that $ \tan(-\theta) = -\tan(\theta) $ and $ \tan(-\frac{\pi}{2}) $ is undefined, but we can use the identity $ \tan(\frac{\pi}{2} - x) = \cot(x) $ to simplify.

Simplify the expression using the identity $ \tan(-\theta - \frac{\pi}{2}) = \cot(\theta) $.

6:14m

Watch next

Master Sum and Difference of Sine & Cosine with a bite sized video explanation from Patrick

Start learning

Sum and Difference Identities practice set

Problem sets built by lead tutors

Expert video explanations

Go to Practice

Struggling with Precalculus?

Expand the expression using the sum & difference identities and simplify.tan(−θ−π2)\tan\left(-\theta-\frac{\pi}{2}\right)tan(−θ−2π)

Watch next

Sum and Difference Identities practice set

Expand the expression using the sum & difference identities and simplify.
$\tan\left(-\theta-\frac{\pi}{2}\right)$