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2 years ago

Prove : (sec θ - tan θ )^2 = 1 - sin θ /1+sin θ

1 answer:

3 0

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

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The G.C.D and L.C.M of three numbers are 3 and 1008 respectively. If two of the numbers are 48 and 72 ,find the least possible v

den301095 [7]

formula= LCM ×GCD of the numbers ÷ The first number

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2 years ago

Your class sells a box of oranges to raise $500 for a field trip.You earn $6.25 for each box of oranges sold.Write and solve an

emmainna [20.7K]

To do this, find the slope, y-intercept, and your goal.

Your goal is 500.
Your slope is 6.25 per orange.
Your y-intercept is 0 (you don't start out with anything.)

So, the inequality is $500 \leq 6.25x$ .

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3 years ago

Read 2 more answers

A random sample of n = 40 observations from a quantitative population produced a mean x = 2.2 and a standard deviation s = 0.29.

zmey [24]

Answer:

$t=\frac{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18$

$df=n-1=40-1=39$

Since is a one side test the p value would be:

$p_v =P(t_{(39)}>2.18)=0.0177$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and then the population mes seems to be higher than 2.1 at 5% of significance

Step-by-step explanation:

Data given and notation

$\bar X=2.2$ represent the sample mean

$s=0.29$ represent the sample standard deviation

$n=40$ sample size

$\mu_o =2.1$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean is higher than 2,1, the system of hypothesis would be:

Null hypothesis: $\mu \leq 2.1$

Alternative hypothesis: $\mu > 2.1$

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{2.2-2.1}{\frac{0.29}{\sqrt{40}}}=2.18$

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n-1=40-1=39$

Since is a one side test the p value would be:

$p_v =P(t_{(39)}>2.18)=0.0177$

Conclusion

7 0

3 years ago

Find the solution for the system of equations -4x-2y=-12 4x +8y=-24

valkas [14]

6y=-36
y=-6
-4x+12=-12
-4x=24
x=-6

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3 years ago

Read 2 more answers

What percent of a dollar is a nickel and a penny? It is ______% of a dollar.

TiliK225 [7]

Answer:

6 %

Step-by-step explana

100 pennies makes a dollar... you just add 5 for the nickel and 1 fo the penny and it makes 6

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2 years ago