Question

Considering only the values of
`\beta` for which
`\sin \beta \tan \beta \sec \beta \cot \beta` is defined, which of the following expressions is equivalent to
`\sin \beta \tan \beta \sec \beta \cot \beta`?

a.
`\sec \beta \cot \beta`
b.
`\tan \beta`
c.
`\cot \beta \tan \beta`
d.
`\tan \beta \csc \beta \sec \beta`

Answer 1

`\huge \boxed{ \boxed{ \red{b) \tan( \beta ) }}}`

Explanation:

to understand this

you need to know about:

• trigonometry
• PEMDAS

given:

• 
  `\sin \beta \tan \beta \sec \beta \cot \beta`

tips and formulas:

• 
  `\tan( \theta) = ( \sin( \theta) )/( \cos( \theta) )`
• 
  `\cot( \theta) = ( \cos( \theta) )/( \sin( \theta) )`
• 
  `\sec( \theta) = (1)/( \cos( \theta) )`

let's solve:

1. 
   `\sf rewrite \: \tan( \beta ) \: as \: ( \sin( \beta ) )/( \cos( \beta ) ) : \\\sin (\beta ) \cdot( \sin( \beta ) )/( \cos( \beta ) ) \cdot \sec (\beta ) \cdot\cot (\beta)`
2. 
   `\sf rewrite \: \sec( \beta ) \: as \: (1 )/( \cos( \beta ) ) : \\\sin (\beta) \cdot( \sin( \beta ) )/( \cos( \beta ) ) \cdot (1)/( \cos( \beta ) ) \cdot\cot (\beta)`
3. 
   `\sf rewrite \: \cot( \beta ) \: as \: ( \cos( \beta ) )/( \sin( \beta ) ) : \\\sin (\beta) \cdot( \sin( \beta ) )/( \cos( \beta ) ) \cdot (1)/( \cos( \beta ) ) \cdot \: ( \cos( \beta ) )/( \sin( \beta ) )`
4. 
   `\sf \: cancel \: sin : \\\sin (\beta) \cdot\frac{ \cancel{\sin( \beta ) }}{ \cos( \beta ) } \cdot (1)/( \cos( \beta ) ) \cdot \: \frac{ \cos( \beta ) }{ \cancel{\sin( \beta ) }} \\ \sin (\beta) \cdot( 1 )/( \cos( \beta ) ) \cdot (1)/( \cos( \beta ) ) \cdot \: \cos( \beta ) \\`
5. 
   `\sf cancel \: cos : \\ \sin (\beta) \cdot( 1)/( \cos( \beta ) ) \cdot \frac{1}{ \cancel{ \cos( \beta )} } \cdot \: \cancel{ \cos( \beta ) } \\ \\ \sin( \beta ) \: \cdot \: ( 1 )/( \cos( \beta ) )`
6. 
   `\sf \: simplify \: multipication : \\ ( \sin( \beta ) )/( \cos( \beta ) )`
7. 
   `\sf use \: ( \sin( \beta ) )/( \cos( \beta ) ) = \tan( \beta ) \: identity : \\ \therefore \: \tan( \beta )`

Answer 2

`\large\boxed{Answer:}`

We will use trigonometric identities to solve this. I will use θ (theta) for the angle.

First of all, we know that cotθ = 1/tanθ. This is a trigonometric identity.

We can replace cotθ in the expression with 1/tanθ.

`sin\theta tan\theta sec\theta (1)/(tan\theta)`

Simplify: 1/tanθ * tanθ = tanθ/tanθ = 1

So now, we have:

`sin\theta sec\theta`

Next, we also know that secθ = 1/cosθ. This is another trigonometric identity.

We can replace secθ with 1/cosθ in our expression.

`sin\theta (1)/(cos\theta)`

Simplify:

`(sin\theta)/(cos\theta)`

Our third trigonometric identity that we will use is tanθ = sinθ/cosθ.

We can replace sinθ/cosθ with tanθ.

Now we have as our final answer:

`\large\boxed{b.\ tan\theta}`

Hope this helps!