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2 years ago

PLEASE HELP!!! TRIG, Theres a picture!!

Mathematics

1 answer:

BartSMP [9]2 years ago

3 0

Answer:

<h2>The answer is option C</h2>

Step-by-step explanation:

$3( { \tan }^{2} \theta - { \sec }^{2} \theta)$

Using trigonometric identities

That's

$1 + { \tan }^{2} \theta = { \sec }^{2} \theta$

Rewrite the expression

That's

$3({ \tan }^{2} \theta - (1 + { \tan }^{2} \theta)) \\$

Simplify

We have

$3({ \tan }^{2} \theta - 1 - { \tan }^{2} \theta) \\ 3({ \tan }^{2} \theta - { \tan }^{2} \theta - 1)$

So we have

3( - 1)

We have the final answer as

Hope this helps you

You might be interested in

If BC = 5, DE = 9, and AB = 4, what is the length of AD?<br><br><br> 7<br> 8<br> 7.2<br> 7.5

steposvetlana [31]

There are two triangles here: ABC and ADF. These triangles are similar, and thus proportional. So, we can set up a proportion between the left side of each triangle and the base of each triangle.

AB / AD = BC / DE

4 / (4 + BD) = 5 / 9

---AD is made up of AB and DB. We know AB, but not BD, so we need to find its value to find the length of AD.

5(4 + BD) = 9 x 4

20 + 5BD = 36

5BD = 16

BD = 3.2

AD = AB + BD

AD = 4 + 3.2

AD = 7.2

Hope this helps!

8 0

3 years ago

Which expression is a fourth root of -1+isqrt3?

aleksklad [387]

Answer:

Step-by-step explanation:

$\sf n^{th}$ roots of a complex number is given by DeMoivre's formula.

$\sf \boxed{\bf r^{\frac{1}{n}}\left[Cos \dfrac{\theta + 2\pi k}{n}+i \ Sin \ \dfrac{\theta+2\pi k}{n}\right]}$

Here, k lies between 0 and (n -1) ; n is the exponent.

$\sf -1 + i\sqrt{3}$

a = -1 and b = √3

$\sf \boxed{r=\sqrt{a^2+b^2}} \ and \ \boxed{\theta = Tan^{-1} \ \dfrac{b}{a}}$

$\sf r = \sqrt{(-1)^2 + 3^2}\\\\ = \sqrt{1+9}\\\\=\sqrt{10}$

$\sf \theta = tan^{-1} \ \dfrac{\sqrt{3}}{-1}\\\\ = tan^{-1} \ (-\sqrt{3})$

$\sf = \dfrac{-\pi }{3}$

n = 4

For k = 0,

$\sf z = \sqrt[4]{10}\left[Cos \ \dfrac{\dfrac{-\pi}{3} +0}{4}+iSin \ \dfrac{\dfrac{-\pi}{3}+0}{4}\right] \\\\\\z= \sqrt[4]{10} \left[Cos \ \dfrac{ -\pi }{12}+iSin \ \dfrac{-\pi}{12}\right]\\\\\\z = \sqrt[4]{10}\left[-Cos \ \dfrac{\pi}{12}-i \ Sin \ \dfrac{\pi}{12}\right]$