Question

Someone show me step for step please!

Answer 1

`\csc \theta \tan \theta - \tan \theta \sin \theta \\\\\\=\frac 1{\sin \theta} \cdot (\sin \theta )/(\cos \theta )- (\sin \theta )/(\cos \theta) \cdot \sin \theta\\\\\\=(1)/(\cos \theta)-(\sin^2 \theta)/(\cos \theta)\\\\\\=(1- \sin^2 \theta)/(\cos \theta)\\\\\\=(\cos^2 \theta)/(\cos \theta)~~~~~~~~~~~~~~;[\sin^2 \theta + \cos^2 \theta =1 ]\\\\\\=\cos \theta`

Answer 2

Let's write out some trigonometric identities that we can use:

• 
  `tan(x)=(sin(x))/(cos(x)) \\`
• 
  `csc(x)=(1)/(sin(x))`
• 
  `sin^2(x)+cos^2(x)=1` ⇒
  `cos^2(x)=1-sin^2(x)`

Now let's try solving them out:

`csc(x)tan(x)-tan(x)sin(x)=(1)/(sin(x))tan(x)-sin(x)tan(x)\\\\ csc(x)tan(x)-tan(x)sin(x)=tan(x)((1)/(sin(x))-sin(x))\\\\ csc(x)tan(x)-tan(x)sin(x)=tan(x)((1)/(sin(x)) -(sin^2(x))/(sin(x)) )\\\\csc(x)tan(x)-tan(x)sin(x)=tan(x)*((1-sin^2(x))/(sin(x)) )\\\\csc(x)tan(x)-tan(x)sin(x)=(sin(x))/(cos(x))*(cos^2(x))/(sin(x)) \\\\csc(x)tan(x)-tan(x)sin(x)=(sin(x))/(sin(x))*(cos^2(x))/(cos(x))\\\\csc(x)tan(x)-tan(x)sin(x)=1 * cos(x)\\\\ csc(x)tan(x)-tan(x)sin(x)=cos(x)`

*I used a different variable, but that doesn't change the answer

Answer: cos(x)

Hope that helps!