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4 years ago

10

How do I prove that (1+sinx)/cosx=cos(-x)/(1+sine(-x))

1 answer:

stira [4]4 years ago

5 0

$\dfrac{1+\sin x}{\cos x}=\dfrac{\cos (-x)}{1+\sin(-x)}\\
\dfrac{1+\sin x}{\cos x}=\dfrac{\cos x}{1-\sin x}\\
\cos^2 x=(1+\sin x)(1-\sin x)\\
\cos^2 x=1-\sin^2 x\\
\sin^2 x+\cos^2 x=1
$

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3 years ago

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3 years ago

The coordinates of ∆ABC are A(-3, 2), B(5, 8) & C(11, 0). Which type of triangle is ∆ABC? Select All that apply.

stepan [7]

Given:

The coordinates of ∆ABC are A(-3, 2), B(5, 8) & C(11, 0).

To find:

The type of the given triangle.

Solution:

Distance formula:

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Using the distance formula, we get

$AB=\sqrt{(5-(-3))^2+(8-2)^2}$

$AB=\sqrt{(8)^2+(6)^2}$

$AB=\sqrt{64+36}$

$AB=\sqrt{100}$

$AB=10$

Similarly,

$BC=\sqrt{(11-5)^2+(0-8)^2}$

$BC=\sqrt{(6)^2+(8)^2}$

$BC=\sqrt{36+64}$

$BC=\sqrt{100}$

$BC=10$

And,

$AC=\sqrt{(11-(-3))^2+(0-2)^2}$

$AC=\sqrt{(14)^2+(-2)^2}$

$AC=\sqrt{196+4}$

$AC=\sqrt{200}$

$AC=10\sqrt{2}$

Two sides of the triangle are equal, i.e., $AB=BC$ . So, the triangle is an isosceles triangle.

Sum of square of two smaller side is

$AB^2+BC^2=10^2+10^2$

$AB^2+BC^2=100+100$

$AB^2+BC^2=200$

$AB^2+BC^2=AC^2$

Using the Pythagoras theorem, we can say that the given triangle is a right triangle.

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3 years ago

AlladinOne [14]

Answer:

okay so I am assuming we are dividing okay so here is how we will do it

here are the top and bottom values written more easily

3^n -3^n-1

3^n+1 +3^n

so first we can cancel out the 3^n values then get left with this

3^n-1

3^n+1

now it depends on the instructions but you can leave it like this or you can subtract their powers to get

3^n-2

Step-by-step explanation:

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3 years ago

What does B.O.D.M.A.S. or B.E.D.M.A.S. mean?

kakasveta [241]

B E D M A S MEANS BRACKETS EXPONENTS DIVISION MULTIPLICATION ADDITION AND SUBTRACTION

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4 years ago

Read 2 more answers