Math / Trigonometry

QuestionQuestion 5 (1 point) $\frac{1}{\cot^2\theta} + \sec\theta \cos\theta$
csc²θ tan²θ sec²θ 1

Studdy Solution

STEP 1

What is this asking? We need to simplify a trigonometric expression involving cotangent, secant, and cosine. Watch out! Remember the fundamental trigonometric identities and be careful not to mix them up!

STEP 2

1. Rewrite in terms of sine and cosine
2. Simplify the expression
3. Apply the Pythagorean identity

STEP 3

Let's rewrite everything in terms of sine and cosine!
This usually makes things much clearer.
We know that $\cot\theta = \frac{\cos\theta}{\sin\theta}$ , so $\cot^2\theta = \frac{\cos^2\theta}{\sin^2\theta}$ .
Therefore, $\frac{1}{\cot^2\theta} = \frac{1}{\frac{\cos^2\theta}{\sin^2\theta}} = \frac{\sin^2\theta}{\cos^2\theta}$ .

STEP 4

Also, we know $\sec\theta = \frac{1}{\cos\theta}$ .
So, $\sec\theta \cdot \cos\theta = \frac{1}{\cos\theta} \cdot \cos\theta = 1$ .
We're multiplying by $\frac{\cos\theta}{\cos\theta}=1$ to add to one.

STEP 5

Now, let's put it all together!
Our original expression becomes: $\frac{1}{\cot^2\theta} + \sec\theta \cdot \cos\theta = \frac{\sin^2\theta}{\cos^2\theta} + 1$

STEP 6

Since $\tan\theta = \frac{\sin\theta}{\cos\theta}$ , we have $\tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta}$ .
So, our expression is now $\tan^2\theta + 1$ .
Getting close!

STEP 7

One of the most important trigonometric identities is $\sin^2\theta + \cos^2\theta = 1$ .
If we divide both sides of this identity by $\cos^2\theta$ , we get: [ \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos

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QuestionQuestion 5 (1 point) $\frac{1}{\cot^2\theta} + \sec\theta \cos\theta$
csc²θ tan²θ sec²θ 1

Studdy Solution

STEP 1

What is this asking? We need to simplify a trigonometric expression involving cotangent, secant, and cosine. Watch out! Remember the fundamental trigonometric identities and be careful not to mix them up!

STEP 2

1. Rewrite in terms of sine and cosine
2. Simplify the expression
3. Apply the Pythagorean identity

STEP 3

STEP 4

Also, we know $\sec\theta = \frac{1}{\cos\theta}$ .
So, $\sec\theta \cdot \cos\theta = \frac{1}{\cos\theta} \cdot \cos\theta = 1$ .
We're multiplying by $\frac{\cos\theta}{\cos\theta}=1$ to add to one.

STEP 5

Now, let's put it all together!
Our original expression becomes: $\frac{1}{\cot^2\theta} + \sec\theta \cdot \cos\theta = \frac{\sin^2\theta}{\cos^2\theta} + 1$

STEP 6

Since $\tan\theta = \frac{\sin\theta}{\cos\theta}$ , we have $\tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta}$ .
So, our expression is now $\tan^2\theta + 1$ .
Getting close!

STEP 7

One of the most important trigonometric identities is $\sin^2\theta + \cos^2\theta = 1$ .
If we divide both sides of this identity by $\cos^2\theta$ , we get: [ \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos

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QuestionQuestion 5 (1 point) 1cot⁡2θ+sec⁡θcos⁡θ\frac{1}{\cot^2\theta} + \sec\theta \cos\thetacot2θ1​+secθcosθcsc2θ tan2θ sec2θ 1

Studdy Solution

STEP 1

What is this asking? We need to simplify a trigonometric expression involving cotangent, secant, and cosine. Watch out! Remember the fundamental trigonometric identities and be careful not to mix them up!

STEP 2

1. Rewrite in terms of sine and cosine2. Simplify the expression3. Apply the Pythagorean identity

STEP 3

STEP 4

STEP 5

Now, let's put it all together!Our original expression becomes: 1cot⁡2θ+sec⁡θ⋅cos⁡θ=sin⁡2θcos⁡2θ+1 \frac{1}{\cot^2\theta} + \sec\theta \cdot \cos\theta = \frac{\sin^2\theta}{\cos^2\theta} + 1 cot2θ1​+secθ⋅cosθ=cos2θsin2θ​+1

STEP 6

Since tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}tanθ=cosθsinθ​, we have tan⁡2θ=sin⁡2θcos⁡2θ\tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta}tan2θ=cos2θsin2θ​.So, our expression is now tan⁡2θ+1\tan^2\theta + 1tan2θ+1.Getting close!

STEP 7

One of the **most important** trigonometric identities is sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1sin2θ+cos2θ=1.If we divide both sides of this identity by cos⁡2θ\cos^2\thetacos2θ, we get: [ \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos

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QuestionQuestion 5 (1 point) $\frac{1}{\cot^2\theta} + \sec\theta \cos\theta$
csc²θ tan²θ sec²θ 1

1. Rewrite in terms of sine and cosine
2. Simplify the expression
3. Apply the Pythagorean identity

Now, let's put it all together!
Our original expression becomes: $\frac{1}{\cot^2\theta} + \sec\theta \cdot \cos\theta = \frac{\sin^2\theta}{\cos^2\theta} + 1$

Since $\tan\theta = \frac{\sin\theta}{\cos\theta}$ , we have $\tan^2\theta = \frac{\sin^2\theta}{\cos^2\theta}$ .
So, our expression is now $\tan^2\theta + 1$ .
Getting close!

One of the most important trigonometric identities is $\sin^2\theta + \cos^2\theta = 1$ .
If we divide both sides of this identity by $\cos^2\theta$ , we get: [ \frac{\sin^2\theta}{\cos^2\theta} + \frac{\cos