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2 years ago

Verify the Identity!!

Mathematics

1 answer:

Elanso [62]2 years ago

6 0

Answer:

Check the proof below

Step-by-step explanation:

try to verify from RHS to LHS

(Tan a + tan b) / (tan a - tan b)

= (sin a / cos a + sin b /cos b) / ( sin a/ cos a - sin b/ cos b)

= (sin a cos b + cos a sin b)/ (sin a cos b - cos a sin b)

= sin (a+b) / sin (a-b)

Fyi

sin(A+B) = sinAcosB+cosAsinB

sin(A-B) = sinAcosB-cosAsinB

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The Office of Student Services at a large western state university maintains information on the study habits of its full-time st

Rudiy27

Answer: The standard error of the mean = 0.5

Step-by-step explanation:

Given : Sample size : n = 144

Standard deviation : $\sigma=6\text{ hours}$

The formula to find the standard error of the mean is given by :-

$S.E.=\dfrac{\sigma}{\sqrt{n}}$

i.e. $S.E.=\dfrac{6}{\sqrt{144}}=0.5$

Hence, the standard error of the mean =0.5

3 0

3 years ago

The first figure is dilated to form the second figure.

ladessa [460]

The scale factor is 2.5

6 0

3 years ago

Read 2 more answers

If x is a positive number greater than 10, what is always true about the solution to the problem to -3 + x ?

alexandr402 [8]

Answer:

The equation answer will always be positive

Step-by-step explanation:

3 0

3 years ago

Would appreciate the help !

aleksandr82 [10.1K]

This is one pathway to prove the identity.

Part 1

$\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{1}{\tan(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\cot(\theta) = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)}{1-\cos(\theta)}-\frac{\cos(\theta)}{\sin(\theta)} = \frac{1}{\sin(\theta)}\\\\\frac{\sin(\theta)*\sin(\theta)}{\sin(\theta)(1-\cos(\theta))}-\frac{\cos(\theta)(1-\cos(\theta))}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\$

Part 2

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

Part 3

$\frac{\sin^2(\theta)+\cos^2(\theta)-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1-\cos(\theta)}{\sin(\theta)(1-\cos(\theta))} = \frac{1}{\sin(\theta)}\\\\\frac{1}{\sin(\theta)} = \frac{1}{\sin(\theta)} \ \ {\checkmark}\\\\$

As the steps above show, the goal is to get both sides be the same identical expression. You should only work with one side to transform it into the other. In this case, the left side transforms while the right side stays fixed the entire time. The general rule is that you should convert the more complicated expression into a simpler form.

We use other previously established or proven trig identities to work through the steps. For example, I used the pythagorean identity $\sin^2(\theta)+\cos^2(\theta) = 1$ in the second to last step. I broke the steps into three parts to hopefully make it more manageable.

3 0

3 years ago

{(4z-7,z-4,z) |z is any real number} z=-5

weeeeeb [17]

The set when z = -5 is {-27, -9, -5}

<h3>How to determine the set elements?</h3>

The set is given as:

{(4z-7,z-4,z) |z is any real number}

When z=-5, we have:

4z - 7 = 4(-5) - 7 = -27

z - 4 = -5 - 4 = -9

z = -5

So, we have:

{(4z-7,z-4,z) |z is any real number} ⇒ {-27, -9, -5}

Hence, the set when z = -5 is {-27, -9, -5}