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Math  /  Trigonometry

Questionsin⁡θ+tan⁡θ1+sec⁡θ\frac{\sin \theta+\tan \theta}{1+\sec \theta}

Studdy Solution

STEP 1

What is this asking? We're asked to simplify a trigonometric expression involving sine, tangent, and secant. Watch out! Remember the basic trigonometric identities and be careful with your algebra!

STEP 2

1. Rewrite everything in terms of sine and cosine.
2. Simplify the numerator.
3. Simplify the whole expression.

STEP 3

Alright, let's **rewrite** everything in terms of sine and cosine!
This is usually a good first step with these kinds of problems.
We know that tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta} and sec⁡θ=1cos⁡θ\sec \theta = \frac{1}{\cos \theta}.
Let's **substitute** these into our expression!

STEP 4

So, our expression becomes: sin⁡θ+sin⁡θcos⁡θ1+1cos⁡θ \frac{\sin \theta + \frac{\sin \theta}{\cos \theta}}{1 + \frac{1}{\cos \theta}} Looking good so far!

STEP 5

Now, let's **focus** on the numerator: sin⁡θ+sin⁡θcos⁡θ\sin \theta + \frac{\sin \theta}{\cos \theta}.
To **add** these terms, we need a common denominator, which is cos⁡θ\cos \theta.
We can **rewrite** sin⁡θ\sin \theta as sin⁡θ⋅cos⁡θcos⁡θ\frac{\sin \theta \cdot \cos \theta}{\cos \theta}.

STEP 6

So, the numerator becomes: sin⁡θ⋅cos⁡θcos⁡θ+sin⁡θcos⁡θ=sin⁡θ⋅cos⁡θ+sin⁡θcos⁡θ \frac{\sin \theta \cdot \cos \theta}{\cos \theta} + \frac{\sin \theta}{\cos \theta} = \frac{\sin \theta \cdot \cos \theta + \sin \theta}{\cos \theta} We can **factor out** sin⁡θ\sin \theta from the top to get: sin⁡θ(cos⁡θ+1)cos⁡θ \frac{\sin \theta (\cos \theta + 1)}{\cos \theta} Awesome!

STEP 7

Let's **look back** at our whole expression.
With our simplified numerator, it now looks like this: sin⁡θ(cos⁡θ+1)cos⁡θ1+1cos⁡θ \frac{\frac{\sin \theta (\cos \theta + 1)}{\cos \theta}}{1 + \frac{1}{\cos \theta}}

STEP 8

We can **rewrite** the denominator 1+1cos⁡θ1 + \frac{1}{\cos \theta} as cos⁡θcos⁡θ+1cos⁡θ=cos⁡θ+1cos⁡θ\frac{\cos \theta}{\cos \theta} + \frac{1}{\cos \theta} = \frac{\cos \theta + 1}{\cos \theta}.
Now, our expression is: sin⁡θ(cos⁡θ+1)cos⁡θcos⁡θ+1cos⁡θ \frac{\frac{\sin \theta (\cos \theta + 1)}{\cos \theta}}{\frac{\cos \theta + 1}{\cos \theta}}

STEP 9

Dividing by a fraction is the same as multiplying by its reciprocal, so we can **rewrite** this as: sin⁡θ(cos⁡θ+1)cos⁡θ⋅cos⁡θcos⁡θ+1 \frac{\sin \theta (\cos \theta + 1)}{\cos \theta} \cdot \frac{\cos \theta}{\cos \theta + 1}

STEP 10

Now, we can **multiply** the fractions: sin⁡θ(cos⁡θ+1)⋅cos⁡θcos⁡θ(cos⁡θ+1) \frac{\sin \theta (\cos \theta + 1) \cdot \cos \theta}{\cos \theta (\cos \theta + 1)} We can **divide** both the numerator and denominator by cos⁡θ(cos⁡θ+1)\cos \theta (\cos \theta + 1) (since it's not zero).
So, we're left with: sin⁡θ \sin \theta Boom!

STEP 11

Our simplified expression is sin⁡θ\sin \theta.

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