Simplify the trigonometric expression tan(2x)/tan(x) using double-angle identities !!

1 answer:

6 0

$\displaystyle\\\text{We use the formula: }~~~\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}\\ ---\\\\\frac{\tan(2x)}{\tan(x)}=\frac{\dfrac{2\tan(x)}{1-\tan^2(x)}}{\tan(x)}=\frac{2\tan(x)}{\tan(x)\Big(1-\tan^2(x)\Big)}=\boxed{\frac{2}{1-\tan^2(x)}}\\\\\\ \text{Correct answer:~~A.}$

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NEED HELP FAST‼️‼️

ch4aika [34]

Answer:

Step-by-step explanation:

A ' = (-2, -3)

B ' = (0, -3)

C ' = (-1, 1)

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Explanation:

To apply an x axis reflection, we simply change the sign of the y coordinate from positive to negative, or vice versa. The x coordinate stays as is.

Algebraically, the reflection rule used can be written as

Applying this rule to the three given points will mean....

Point A = (-2, 3) becomes A ' = (-2, -3)

Point B = (0, 3) becomes B ' = (0, -3)

Point C = (-1, -1) becomes C ' = (-1, 1)

The diagram is provided below.

Side note: Any points on the x axis will stay where they are. That isn't the case here, but its for any future problem where it may come up. This only applies to x axis reflections.

4 0

3 years ago

Find the area of each regular polygon. Round your answer to the nearest tenth if necessary.

tatuchka [14]

*I am assuming that the hexagons in all questions are regular and the triangle in (24) is equilateral*

(21)

Area of a Regular Hexagon: $\frac{3\sqrt{3}}{2}(side)^{2} = \frac{3\sqrt{3}}{2}*(\frac{20\sqrt{3} }{3} )^{2} =200\sqrt{3}$ square units

(22)

Similar to (21)

Area = $216\sqrt{3}$ square units

(23)

For this case, we will have to consider the relation between the side and inradius of the hexagon. Since, a hexagon is basically a combination of six equilateral triangles, the inradius of the hexagon is basically the altitude of one of the six equilateral triangles. The relation between altitude of an equilateral triangle and its side is given by:

$altitude=\frac{\sqrt{3}}{2}*side$