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Math  /  Trigonometry

QuestionVerify the identity using the fundamental trigonometric identities. tan⁡(θ)+cot⁡(θ)=sec⁡(θ)csc⁡(θ)\tan (\theta)+\cot (\theta)=\sec (\theta) \csc (\theta)
Use Reciprocal Identities to rewrite the expression in terms of sine and cosin tan⁡(θ)+cot⁡(θ)=sin⁡(θ)cos⁡(θ)+□sin⁡(θ)=□cos⁡(θ)sin⁡(θ)\begin{aligned} \tan (\theta)+\cot (\theta) & =\frac{\sin (\theta)}{\cos (\theta)}+\frac{\square}{\sin (\theta)} \\ & =\frac{\square}{\cos (\theta) \sin (\theta)} \end{aligned}
Use a Pythagorean Identity to simplify the numerator of the expression. =□cos⁡(θ)sin⁡(θ)=\frac{\square}{\cos (\theta) \sin (\theta)}
Use Reciprocal Identities again to simplify. =□=\square Submit Answer
15. [1/1 Points] DETAILS MY NOTES SPRECALC8 7.1.042.

Studdy Solution

STEP 1

What is this asking? We need to prove that tan⁡(θ)+cot⁡(θ)\tan(\theta) + \cot(\theta) is exactly the same as sec⁡(θ)csc⁡(θ)\sec(\theta)\csc(\theta) using some cool trig tricks! Watch out! Don't mix up your reciprocal and Pythagorean identities!
Also, be super careful with your signs when simplifying.

STEP 2

1. Rewrite in terms of Sine and Cosine
2. Common Denominator
3. Pythagorean Identity
4. Reciprocal Identities

STEP 3

We know that tan⁡(θ)\tan(\theta) is the same as sin⁡(θ)cos⁡(θ)\frac{\sin(\theta)}{\cos(\theta)} and cot⁡(θ)\cot(\theta) is the same as cos⁡(θ)sin⁡(θ)\frac{\cos(\theta)}{\sin(\theta)}.
So, let's **rewrite** our starting equation: tan⁡(θ)+cot⁡(θ)=sin⁡(θ)cos⁡(θ)+cos⁡(θ)sin⁡(θ) \tan(\theta) + \cot(\theta) = \frac{\sin(\theta)}{\cos(\theta)} + \frac{\cos(\theta)}{\sin(\theta)} This makes things easier to work with!

STEP 4

To add these two fractions, we need a **common denominator**.
Let's multiply the first fraction by sin⁡(θ)sin⁡(θ)\frac{\sin(\theta)}{\sin(\theta)} (which is just a fancy way of multiplying by **one**) and the second fraction by cos⁡(θ)cos⁡(θ)\frac{\cos(\theta)}{\cos(\theta)} (also a fancy **one**!). sin⁡(θ)cos⁡(θ)⋅sin⁡(θ)sin⁡(θ)+cos⁡(θ)sin⁡(θ)⋅cos⁡(θ)cos⁡(θ) \frac{\sin(\theta)}{\cos(\theta)} \cdot \frac{\sin(\theta)}{\sin(\theta)} + \frac{\cos(\theta)}{\sin(\theta)} \cdot \frac{\cos(\theta)}{\cos(\theta)}

STEP 5

This gives us: sin⁡2(θ)cos⁡(θ)sin⁡(θ)+cos⁡2(θ)cos⁡(θ)sin⁡(θ) \frac{\sin^2(\theta)}{\cos(\theta)\sin(\theta)} + \frac{\cos^2(\theta)}{\cos(\theta)\sin(\theta)} Now we can **add** the fractions: sin⁡2(θ)+cos⁡2(θ)cos⁡(θ)sin⁡(θ) \frac{\sin^2(\theta) + \cos^2(\theta)}{\cos(\theta)\sin(\theta)}

STEP 6

Remember the **Pythagorean identity**: sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1?
Let's **use** it!
We can **replace** the numerator with a nice and simple **1**: 1cos⁡(θ)sin⁡(θ) \frac{1}{\cos(\theta)\sin(\theta)}

STEP 7

We know that 1cos⁡(θ)\frac{1}{\cos(\theta)} is sec⁡(θ)\sec(\theta) and 1sin⁡(θ)\frac{1}{\sin(\theta)} is csc⁡(θ)\csc(\theta).
So, we can **rewrite** our expression as: 1cos⁡(θ)⋅1sin⁡(θ)=sec⁡(θ)csc⁡(θ) \frac{1}{\cos(\theta)} \cdot \frac{1}{\sin(\theta)} = \sec(\theta)\csc(\theta)

STEP 8

We've shown that tan⁡(θ)+cot⁡(θ)=sec⁡(θ)csc⁡(θ)\tan(\theta) + \cot(\theta) = \sec(\theta)\csc(\theta)!
We've successfully **verified** the identity!

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