In order to solve this problem, one must envision a right triangle. A diagram used to represent the imagined right triangle is included at the bottom of this explanation. Please note that each side is named with respect to the angle is it across from.

Right angle trigonometry is composed of a sequence of ratios that relate the sides and angles of a right triangle. These ratios are as follows,

$sin(\theta)=\frac{opposite}{hypotenuse}\\\\cos(\theta)=\frac{adjacent}{hypotenuse}\\\\tan(\theta)=\frac{opposite}{adjacent}$

One is given the following equation,

$\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0$

As per the attached reference image, one can state the following, using the right angle trigonometric ratios,

$sin(A)=\frac{a}{c}\\\\sin(B)=\frac{b}{c}\\\\cos(A)=\frac{b}{c}\\\\cos(B)=\frac{a}{c}$

Substitute these values into the given equation. Then simplify the equation to prove the idenity,

$\frac{sin(A)+sin(B)}{cos(A) +cos(B)}+\frac{cos(A)-cos(B)}{sin(A)-sin(B)}=0$

$\frac{\frac{a}{c}+\frac{b}{c}}{\frac{b}{c}+\frac{a}{c}}+\frac{\frac{b}{c}-\frac{a}{c}}{\frac{a}{c}-\frac{b}{c}}=0$

$\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}$

Remember, any number over itself equals one, this holds true even for fractions with fractions in the numerator (value on top of the fraction bar) and denominator (value under the fraction bar).

$\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{b-a}{c}}{\frac{a-b}{c}}$

$\frac{\frac{a+b}{c}}{\frac{a+b}{c}}+\frac{\frac{-(a-b)}{c}}{\frac{a-b}{c}}$

$1+(-1)=0$

$1-1=0$

$0=0$

7 0

3 years ago

If you spend $448.37 on a car each month, what is your new subtotal of take-home pay after expenses? $480.29 $629.75 $779.21 $92

Black_prince [1.1K]

I just had the question and the answer is $629.75

8 0

3 years ago

Read 2 more answers

On the planet of Mercury, 4-year-olds average 3 hours a day unsupervised. Most of the unsupervised children live in rural areas,

erastova [34]

Answer:

The distribution of X is normal

$P(X < 1.4) = 0.12654$

$P(X > 3.5) = 0.36051$

Step-by-step explanation:

From the question we are told that

The population mean is $\mu = 3 \ hours$

The standard deviation is $\sigma = 1.4 \ hours$

Generally given from the question that the amount of time spent alone by the population size is normally distributed then then the distribution of X (i.e the amount of time spent by the sample size (the one Mercurian)) will be normally distributed

Generally the probability that the child spend less than one hour in a day is mathematically represented as

$P(X < 1.4) = P(\frac{X - \mu}{\sigma} < \frac{1.4 - \mu}{\sigma} )$