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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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2 u + 12 < 23 0,1 d + 8 > 0 3 - 4 r > 7 13 < 2 z + 3 6 ≥ 9 - 0, 15 i

BartSMP [9]

Answer:

Part 1) $u< 5.5$

Part 2) $d > -80$

Part 3) $r < -1$

Part 4) $z>5$

Part 5) $i\geq 20$

Step-by-step explanation:

The question is

Solve each inequality for the indicated variable

Part 1) we have

$2u+12$

subtract 12 both sides

$2u< 23-12\\2u$

Divide by 2 both sides

$u< 5.5$

The solution is the interval (-∞,5.5)

In a number line the solution is the shaded area at left of u=5.5 (open circle)

The number 5.5 is not included in the solution

Part 2) we have

$0.1d+8 >0$

subtract 8 both sides

$0.1d > -8$

Divide by 0.1 both sides

$d > -80$

The solution is the interval (-80,∞)

In a number line the solution is the shaded area at right of d=-80 (open circle)

The number -80 is not included in the solution

Part 3) we have

$3-4r> 7$

Subtract 3 both sides

$-4r>7-3\\-4r>4$

Divide by -4 both sides

Remember that, when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

$r < -1$

The solution is the interval (-∞,-1)

In a number line the solution is the shaded area at left of r=-1.1 (open circle)

The number -1.1 is not included in the solution

Part 4) we have

$13< 2z+3$

Subtract 3 both sides

$13-3$

Divide by 2 both sides

$5$

Rewrite

$z>5$

The solution is the interval (5,∞)

In a number line the solution is the shaded area at right of z=5 (open circle)

The number 5 is not included in the solution

Part 5) we have

$6\geq 9-0.15i$

Subtract 9 both sides

$6-9\geq -0.15i\\-3\geq -0.15i$

Divide by -0.15 both sides

Remember that, when you multiply or divide both sides of an inequality by a negative number, you must reverse the inequality symbol

$20\leq i$

Rewrite

$i\geq 20$

The solution is the interval [20,∞)

In a number line the solution is the shaded area at right of i=20 (closed circle)

The number 20 is included in the solution

6 0

3 years ago

A sphere with a radius of 6cm has the same volume as a cylinder with a height of 4.5cm. What is the radius of the cylinder

yawa3891 [41]

Short answer: r = 8
Remark
The easiest way to do this is to solve the sphere's volume in terms of pi. When you do this, you can equate that to the formula for a cylinder and cancel the pi values.

Step One
Find the volume of the sphere.

Givens
r = 6 cm

Formula
V = (4/3) pi r^3

Sub and Solve
V = 4/3 pi * 6^3
V = 288 * pi

Step two
Find the radius of the cylinder

Givens
V = 288* pi cm^3
h = 4.5 cm

Formula
V = pi r^h

Sub and solve
288 pi cm^3 = pi r^2 * 4.5 Divide both sides by pi
288 cm^3 = 4.5 r^2 Divide both sides by 4.5
388 / 4.5 = r^2
64 = r^2 Take the square root of both sides.
r = square root( 64)
r = 8 <<<<< Answer

4 0

2 years ago

14=2y/5 What’s the answer

shutvik [7]

The answer is 35 ok ?

7 0

2 years ago

What is 900cm/h = cm/min? ( with an explanation not just the answer)

vladimir2022 [97]

I hope that helped :)

6 0

3 years ago

Find the equation of the plane that is parallel to the vectors left angle 3 comma 0 comma 3 right angle and left angle 0 comma 1

Sholpan [36]

Answer:

$x + 3y -z - 3 = 0$

Step-by-step explanation:

We have to find the equation of plane that is parallel to the vectors

$\langle 3,0,3\rangle, \langle0,1,3\rangle$

The plane also passes through the point (2,0,-1).

Hence, the equation of plane s given by:

$\displaystyle\left[\begin{array}{ccc}x-2&y-0&z+1\\3&0&3\\0&1&3\end{array}\right]\\\\=(x-2)(0-3) - (y-0)(9-0) + (z+1)(3-0)\\=-3(x-2)-9y+3(z+1)\\\Rightarrow -3x + 6 - 9y + 3z + 3 = 0\\\Rightarrow 3x + 9y -3z -9 = 0\\\Rightarrow x + 3y -z - 3 = 0$

It is the required equation of plane.

4 0

3 years ago