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Math  /  Trigonometry

QuestionEstablish the identity. sec⁡θ+tan⁡θtan⁡θ−sec⁡θ+tan⁡θsec⁡θ=cos⁡θcot⁡θ\frac{\sec \theta+\tan \theta}{\tan \theta}-\frac{\sec \theta+\tan \theta}{\sec \theta}=\cos \theta \cot \theta

Studdy Solution

STEP 1

What is this asking? We need to prove that two trigonometric expressions are equal to each other.
We're going to manipulate one side of the equation to make it look exactly like the other side! Watch out! Remember your basic trig identities, and don't accidentally flip them around!
Also, be careful with your algebra when combining fractions.

STEP 2

1. Rewrite in terms of sine and cosine
2. Combine the fractions
3. Simplify the numerator
4. Simplify the expression

STEP 3

Let's **rewrite** everything in terms of sine and cosine!
This usually makes things much clearer.
Remember, sec⁡θ=1cos⁡θ\sec \theta = \frac{1}{\cos \theta} and tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}.
So, our starting expression becomes:
1cos⁡θ+sin⁡θcos⁡θsin⁡θcos⁡θ−1cos⁡θ+sin⁡θcos⁡θ1cos⁡θ \frac{\frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} - \frac{\frac{1}{\cos \theta} + \frac{\sin \theta}{\cos \theta}}{\frac{1}{\cos \theta}}

STEP 4

Now, let's **combine** those fractions in the numerators.
Since they already have a common denominator of cos⁡θ\cos \theta, this is easy!
1+sin⁡θcos⁡θsin⁡θcos⁡θ−1+sin⁡θcos⁡θ1cos⁡θ \frac{\frac{1 + \sin \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}} - \frac{\frac{1 + \sin \theta}{\cos \theta}}{\frac{1}{\cos \theta}}

STEP 5

Next, let's **simplify** by remembering how to divide fractions: we multiply by the reciprocal of the denominator!
1+sin⁡θcos⁡θ⋅cos⁡θsin⁡θ−1+sin⁡θcos⁡θ⋅cos⁡θ1 \frac{1 + \sin \theta}{\cos \theta} \cdot \frac{\cos \theta}{\sin \theta} - \frac{1 + \sin \theta}{\cos \theta} \cdot \frac{\cos \theta}{1}

STEP 6

Look at that, we can **multiply** those cos⁡θ\cos \theta terms in both parts of the expression.
Remember, anything divided by itself is **one**, so cos⁡θcos⁡θ=1\frac{\cos \theta}{\cos \theta} = 1.
1+sin⁡θsin⁡θ⋅1−1+sin⁡θ1⋅1 \frac{1 + \sin \theta}{\sin \theta} \cdot 1 - \frac{1 + \sin \theta}{1} \cdot 1

STEP 7

This **simplifies** to:
1+sin⁡θsin⁡θ−(1+sin⁡θ) \frac{1 + \sin \theta}{\sin \theta} - (1 + \sin \theta)

STEP 8

Now, let's **rewrite** the first fraction and then **distribute** the minus sign in the second part:
1sin⁡θ+sin⁡θsin⁡θ−1−sin⁡θ \frac{1}{\sin \theta} + \frac{\sin \theta}{\sin \theta} - 1 - \sin \theta

STEP 9

Since anything divided by itself is **one**, this **simplifies** to:
1sin⁡θ+1−1−sin⁡θ \frac{1}{\sin \theta} + 1 - 1 - \sin \theta

STEP 10

The **ones add to zero**, leaving us with:
1sin⁡θ−sin⁡θ \frac{1}{\sin \theta} - \sin \theta

STEP 11

Let's **rewrite** sin⁡θ\sin \theta as sin⁡2θsin⁡θ\frac{\sin^2 \theta}{\sin \theta} so we can **combine** the fractions:
1sin⁡θ−sin⁡2θsin⁡θ=1−sin⁡2θsin⁡θ \frac{1}{\sin \theta} - \frac{\sin^2 \theta}{\sin \theta} = \frac{1 - \sin^2 \theta}{\sin \theta}

STEP 12

Remember the **Pythagorean identity**: sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1.
If we subtract sin⁡2θ\sin^2 \theta from both sides, we get 1−sin⁡2θ=cos⁡2θ1 - \sin^2 \theta = \cos^2 \theta.
Let's **use** this!
cos⁡2θsin⁡θ \frac{\cos^2 \theta}{\sin \theta}

STEP 13

We can **rewrite** this as:
cos⁡θ⋅cos⁡θsin⁡θ \cos \theta \cdot \frac{\cos \theta}{\sin \theta}

STEP 14

And since cos⁡θsin⁡θ=cot⁡θ\frac{\cos \theta}{\sin \theta} = \cot \theta, we **finally** get:
cos⁡θcot⁡θ \cos \theta \cot \theta

STEP 15

We've successfully shown that sec⁡θ+tan⁡θtan⁡θ−sec⁡θ+tan⁡θsec⁡θ\frac{\sec \theta+\tan \theta}{\tan \theta}-\frac{\sec \theta+\tan \theta}{\sec \theta} simplifies to cos⁡θcot⁡θ\cos \theta \cot \theta!

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