Get everything together into one fraction by finding the LCD and doing the math. The LCD is sin(x) cos(x). Multiplying that in to each term looks like this:

$[sin(x)cos(x)]\frac{sin(x)}{cos(x)}+[sin(x)cos(x)]\frac{cos(x)}{sin(x)} =?$

In the first term, the cos(x)'s cancel out, and in the second term the sin(x)'s cancel out, leaving:

$\frac{sin^2(x)}{sin(x)cos(x)}+\frac{cos^2(x)}{sin(x)cos(x)}=?$

Put everything over the common denominator now:

$\frac{sin^2(x)+cos^2(x)}{sin(x)cos(x)}=?$

Since $sin^2(x)+cos^2(x)=1$ , we will make that substitution:

$\frac{1}{sin(x)cos(x)}$

We could separate that fraction into 2:

$\frac{1}{sin(x)}$ × $\frac{1}{cos(x)}$

$\frac{1}{sin(x)}=csc(x)$ and $\frac{1}{cos(x)}=sec(x)$

Therefore, the simplification is

sec(x)csc(x)

5 0

3 years ago

4.) The time X (minutes) for a lab assistant to prepare the equipment for a certain experiment is believed to have a uniform dis

Jet001 [13]

Answer:

$P(35-2 < X 35+2) = P(33< X< 37)= P(X$

And using the cumulative distribution function we got:

$P(35-2 < X 35+2) = P(33< X< 37)= P(X$

The probability that preparation is within 2 minutes of the mean time is 0.134

Step-by-step explanation:

For this case we define the following random variable X= (minutes) for a lab assistant to prepare the equipment for a certain experiment , and the distribution for X is given by:

$X \sim Unif (a= 20, b =50)$