Question

Someone show me step for step please!

Answer 1

$\csc \theta \tan \theta - \tan \theta \sin \theta \\\\\\=\dfrac 1{\sin \theta} \cdot \dfrac{\sin \theta }{\cos \theta }- \dfrac{\sin \theta }{\cos \theta} \cdot \sin \theta\\\\\\=\dfrac{1}{\cos \theta}-\dfrac{\sin^2 \theta}{\cos \theta}\\\\\\=\dfrac{1- \sin^2 \theta}{\cos \theta}\\\\\\=\dfrac{\cos^2 \theta}{\cos \theta}~~~~~~~~~~~~~~;[\sin^2 \theta + \cos^2 \theta =1 ]\\\\\\=\cos \theta$

Answer 2

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

• $tan(x)=\frac{sin(x)}{cos(x)}$
• $csc(x)=\frac{1}{sin(x)}$
• $sin^2(x)+cos^2(x)=1$ ⇒ $cos^2(x)=1-sin^2(x)$

Now let's try solving them out:

  $csc(x)tan(x)-tan(x)sin(x)=\frac{1}{sin(x)}tan(x)-sin(x)tan(x)\\\\ csc(x)tan(x)-tan(x)sin(x)=tan(x)(\frac{1}{sin(x)}-sin(x))\\\\ csc(x)tan(x)-tan(x)sin(x)=tan(x)(\frac{1}{sin(x)} -\frac{sin^2(x)}{sin(x)} )\\\\csc(x)tan(x)-tan(x)sin(x)=tan(x)*(\frac{1-sin^2(x)}{sin(x)} )\\\\csc(x)tan(x)-tan(x)sin(x)=\frac{sin(x)}{cos(x)}*\frac{cos^2(x)}{sin(x)} \\\\csc(x)tan(x)-tan(x)sin(x)=\frac{sin(x)}{sin(x)}*\frac{cos^2(x)}{cos(x)}\\\\csc(x)tan(x)-tan(x)sin(x)=1 * cos(x)\\\\ csc(x)tan(x)-tan(x)sin(x)=cos(x)$

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

Answer: cos(x)

Hope that helps!