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3 years ago

Prove that 1+cosA/sinA + sinA/1+cosA=2cosecA

1 answer:

5 0

$\frac{1+cos\alpha}{sin\alpha}+\frac{sin\alpha}{1+cos\alpha}=2cosec\alpha\\\\L=\frac{(1+cos\alpha)(1+cos\alpha)+sin\alpha\cdot sin\alpha}{sin\alpha(1+cos\alpha)}=\frac{1+2cos\alpha+cos^2\alpha+sin^2\alpha}{sin\alpha(1+cos\alpha)}\\\\=\frac{1+2cos\alpha+1}{sin\alpha(1+cos\alpha)}=\frac{2+2cos\alpha}{sin\alpha(1+cos\alpha)}=\frac{2(1+cos\alpha)}{sin\alpha(1+cos\alpha)}\\\\=\frac{2}{sin\alpha}=2\cdot\frac{1}{sin\alpha}=2cosec\alpha=R$

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C) The opposite of the least 2-digit positive integer is what

Dmitry_Shevchenko [17]

That would be -10...

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3 years ago

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I need help with 11,12,13

liraira [26]

The ratios should be,
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3 years ago

a museum opened at 8:00 am. in the first hour 350 people purchased admission tickets. in the 2nd hour 20% more people purchased

Levart [38]

Okay, so we already know that 350 people purchased tickets in hour 1:

17.50(350 + ?) = ?

The question states that in the 2nd hour, there was 20% more people. So basically, we solve for 120% of the people from the 1st hour:

17.50(350 + 1.2(350)) = ?

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17.50(350 + 420) = ?

Add:

17.50(770) = ?

Multiply:

17.50(770) = 13475

$13,475 was the total amount of money.

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3 years ago

5milimeter7centimeter+1milimiter=

Marrrta [24]

7milimeters7centumeters

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4 years ago

(08.07 HC)

andreev551 [17]

Answer:

$\textsf{A)} \quad x=-2, \:\:x=\dfrac{5}{2}$

$\textsf{B)} \quad \left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)$

C) See attachment.

Step-by-step explanation:

Given function:

$f(x)=2x^2-x-10$

To factor a <u>quadratic</u> in the form $ax^2+bx+c$ <em> , </em>find two numbers that multiply to $ac$ and sum to $b$ :

$\implies ac=2 \cdot -10=-20$

$\implies b=-1$

Therefore, the two numbers are -5 and 4.

Rewrite $b$ as the sum of these two numbers:

$\implies f(x)=2x^2-5x+4x-10$

Factor the first two terms and the last two terms separately:

$\implies f(x)=x(2x-5)+2(2x-5)$

Factor out the common term (2x - 5):

$\implies f(x)=(x+2)(2x-5)$

The x-intercepts are when the curve crosses the x-axis, so when y = 0:

$\implies (x+2)(2x-5)=0$

Therefore:

$\implies (x+2)=0 \implies x=-2$

$\implies (2x-5)=0 \implies x=\dfrac{5}{2}$

So the x-intercepts are:

$x=-2, \:\:x=\dfrac{5}{2}$

The x-value of the vertex is:

$\implies x=\dfrac{-b}{2a}$

Therefore, the x-value of the vertex of the given function is:

$\implies x=\dfrac{-(-1)}{2(2)}=\dfrac{1}{4}$

To find the y-value of the vertex, substitute the found value of x into the function:

$\implies f\left(\dfrac{1}{4}\right)=2\left(\dfrac{1}{4}\right)^2-\left(\dfrac{1}{4}\right)-10=-\dfrac{81}{8}$

Therefore, the vertex of the function is:

$\left(\dfrac{1}{4},-\dfrac{81}{8}\right)=(0.25,-10.125)$

Plot the x-intercepts found in Part A.

Plot the vertex found in Part B.

As the <u>leading coefficient</u> of the function is positive, the parabola will open upwards. This is confirmed as the vertex is a minimum point.

The axis of symmetry is the <u>x-value</u> of the <u>vertex</u>. Draw a line at x = ¹/₄ and use this to ensure the drawing of the parabola is <u>symmetrical</u>.

Draw a upwards opening parabola that has a minimum point at the vertex and that passes through the x-intercepts (see attachment).

5 0

2 years ago