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

$\large\boxed{Answer:}$

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.

$sin\theta tan\theta sec\theta \frac{1}{tan\theta}$

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.

$sin\theta \frac{1}{cos\theta}$

Simplify:

$\frac{sin\theta}{cos\theta}$

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!

Answer 2

Answer:

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

Step-by-step explanation:

to understand this

you need to know about:

• trigonometry
• PEMDAS

given:

• $\sin \beta \tan \beta \sec \beta \cot \beta$

tips and formulas:

• $\tan( \theta) = \dfrac{ \sin( \theta) }{ \cos( \theta) }$
• $\cot( \theta) = \dfrac{ \cos( \theta) }{ \sin( \theta) }$
• $\sec( \theta) = \dfrac{1}{ \cos( \theta) }$

let's solve:

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