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Lubov Fominskaja [6]
2 years ago
13

Prove :.........

3D%20%20%5Cfrac%7Bsinx%7D%7B1%20-%20cosx%7D%20" id="TexFormula1" title=" \sqrt{ \frac{1 + cosx}{1 - cosx} } = \frac{sinx}{1 - cosx} " alt=" \sqrt{ \frac{1 + cosx}{1 - cosx} } = \frac{sinx}{1 - cosx} " align="absmiddle" class="latex-formula">
​
Mathematics
1 answer:
krok68 [10]2 years ago
6 0

This is generally not true. Let x = 3π/2. Then

cos(3π/2) = 0

sin(3π/2) = -1

and

√((1 + cos(x)) / (1 - cos(x))) = √(1 / 1) = 1

while

sin(x) / (1 - cos(x)) = -1 / 1 = -1

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Answer:  6\sqrt{3}

======================================================

Explanation:

Method 1

We can use the pythagorean theorem to find x.

a^2+b^2 = c^2\\\\6^2+x^2 = 12^2\\\\36+x^2 = 144\\\\x^2 = 144-36\\\\x^2 = 108\\\\x = \sqrt{108}\\\\x = \sqrt{36*3}\\\\x = \sqrt{36}*\sqrt{3}\\\\x = 6\sqrt{3}\\\\

-----------------------------------

Method 2

Use the sine ratio to find x. You'll need a reference sheet or the unit circle, or simply memorize that sin(60) = sqrt(3)/2

\sin(\text{angle}) = \frac{\text{opposite}}{\text{hypotenuse}}\\\\\sin(60^{\circ}) = \frac{x}{12}\\\\\frac{\sqrt{3}}{2} = \frac{x}{12}\\\\x = 12*\frac{\sqrt{3}}{2}\\\\x = 6\sqrt{3}\\\\

-----------------------------------

Method 3

Similar to the previous method, but we'll use tangent this time.

Use a reference sheet, unit circle, or memorize that tan(60) = sqrt(3)

\tan(\text{angle}) = \frac{\text{opposite}}{\text{adjacent}}\\\\\tan(60^{\circ}) = \frac{x}{6}\\\\\sqrt{3} = \frac{x}{6}\\\\x = 6\sqrt{3}\\\\

-----------------------------------

Method 4

This is a 30-60-90 triangle. In other words, the angles are 30 degrees, 60 degrees, and 90 degrees.

Because of this special type of triangle, we know that the long leg is exactly sqrt(3) times that of the short leg.

\text{long leg} = (\text{short leg})*\sqrt{3}\\\\x = 6\sqrt{3}\\\\

The short leg is always opposite the smallest angle (30 degrees).

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