Question

Cleft%28x%5Cright%29%7D%7B%5Ccsc%5Cleft%28x%5Cright%29%5Ccos%5E%7B2%7D%5Cleft%28x%5Cright%29%7D" id="TexFormula1" title="\frac{\sec\left(x\right)}{\cos\left(x\right)}-\frac{\sin\left(x\right)}{\csc\left(x\right)\cos^{2}\left(x\right)}" alt="\frac{\sec\left(x\right)}{\cos\left(x\right)}-\frac{\sin\left(x\right)}{\csc\left(x\right)\cos^{2}\left(x\right)}" align="absmiddle" class="latex-formula">Use the basic identities to change the expression to one involving only sines and cosines. Then simplify to a basic trig function.

Answer 1

Answer:

1

Step-by-step explanation:

First, convert all the secants and cosecants to cosine and sine, respectively. Recall that $csc(x)=1/sin(x)$ and $sec(x)=1/cos(x)$.

Thus:

$\frac{sec(x)}{cos(x)} -\frac{sin(x)}{csc(x)cos^2(x)}$

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

Let's do the first part first: (Recall how to divide fractions)

$\frac{\frac{1}{cos(x)} }{cos(x)}=\frac{1}{cos(x)} \cdot \frac{1}{cos(x)}=\frac{1}{cos^2(x)}$

For the second term:

$\frac{sin(x)}{\frac{cos^2(x)}{sin(x)} } =\frac{sin(x)}{1} \cdot\frac{sin(x)}{cos^2(x)}=\frac{sin^2(x)}{cos^2(x)}$

So, all together: (same denominator; combine terms)

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

Note the numerator; it can be derived from the Pythagorean Identity:

$sin^2(x)+cos^2(x)=1; cos^2(x)=1-sin^2(x)$

Thus, we can substitute the numerator:

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

Everything simplifies to 1.