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3 years ago

Sec^6x (secxtanx) - sec^4x (secxtanx) = sec^5x tan^3x

1 answer:

3 0

Let E the expression of the left side of the equation:

$E=\sec^6(x) (\sec (x)\tan (x)) - \sec^4(x) (\sec (x)\tan (x))\\\\
E=\sec^7(x)\tan(x)-\sec^5(x)\tan(x)\\\\
E=\sec^5(x)\tan(x)(\sec^2(x)-1)$

We must know that $\tan^2(x)+1=\sec^2(x)\iff \sec^2(x)-1=\tan^2(x)$ . So, using in the equation above:

$E=\sec^5(x)\tan(x)(\underbrace{\sec^2(x)-1}_{\tan^2(x)})\\\\
E=\sec^5(x)\tan(x)(\tan^2(x))\\\\
\boxed{E=\sec^5(x)\tan^3(x)}$

Then, the equation is true.

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If the second number is subtracted from the sum of the first number and 2 times the third number, the result is 1. The thrid num

weeeeeb [17]

Answer:

Step-by-step explanation:

$x,\ y,\ z-\text{three numbers}\\\\\left\{\begin{array}{ccc}(x+2z)-y=1&(1)\\z+2x=3&(2)\\x+3y+z=18&(3)\end{array}\right\\\\(2)\\z+2x=3\qquad\text{subtract}\ 2x\ \text{from both sides}\\z=3-2x\qquad(*)\\\\\text{Substitute}\ (*)\ \text{to (1) and (3)}\\\\\left\{\begin{array}{ccc}x+2(3-2x)-y=1&\text{use the distributive property}\\x+3y+(3-2x)=18\end{array}\right$

$\left\{\begin{array}{ccc}x+(2)(3)+(2)(-2x)-y=1\\x+3y+3-2x=18&\text{subtract 3 from both sides}\end{array}\right\\\left\{\begin{array}{ccc}x+6-4x-y=1&\text{subtract 6 from both sides}\\(x-2x)+3y=15\end{array}\right\\\left\{\begin{array}{ccc}(x-4x)-y=-5\\-x+3y=15\end{array}\right\\\left\{\begin{array}{ccc}-3x-y=-5&\text{multiply both sides by 3}\\-x+3y=15\end{array}\right$

$\underline{+\left\{\begin{array}{ccc}-9x-3y=-15\\-x+3y=15\end{array}\right}\qquad\text{add all sides of the equations}\\.\qquad-10x=0\qquad\text{divide both sides by (-10)}\\.\qquad\boxed{x=0}\\\\\text{Put it to the second equation:}\\-0+3y=15\\3y=15\qquad\text{divide both sides by 3}\\\boxed{y=5}\\\\\text{Put the value of}\ x\ \text{to}\ (*):\\\\z=3-2(0)\\\boxed{z=3}$

7 0

3 years ago

Which of the statements about the following quadratic equation is true?

Sedaia [141]

After manipulating above equation we get, 2x^2 - 7x - 8= 0. Discriminant= b^2 - 4ac = (-7)^2 - 4(2)(-8) = 113>0. So there are 2 real roots :)

3 0

3 years ago

Read 2 more answers

Tell whether the lines through the given points are parallel, perpendicular, or neither.

antoniya [11.8K]

Answer:

The Slope of Line 1 is -> $-\frac{2}{9}$

The Slope of Line 2 is -> $-\frac{9}{2}$

These two lines are - > Perpendicular

Step-by-step explanation:

To find the slope of line one, you follow these steps:

First Coordinate 10 -> x1, 5 -> y1

Second Coordinate -8 -> x2, 9 -> y2

Then insert the assigned values into this equation: $\frac{y2-y1}{x2-x1}$ -> $\frac{9-5}{-8-10}$ which equals $\frac{4}{-18}$ which simplifies to $-\frac{2}{9}$

Then do the same process to find the slop of line 2:

First Coord: 2 -> x1, -4 -> y1

Second Coord: 11 -> x2, -6 -> y2

Then insert the values into this equation: $\frac{y2-y1}{x2-x1}$ -> $\frac{11-2}{-6-(-4)}$ which equals $\frac{9}{-2}$ .

We know the lines are perpendicular by:

a. Typing the equations of the lines into a graphing calculator and observing the graph

b. Looking at the slopes. By looking at the slopes, we see that the slope of line 1 is the reciprocal of line 2, and vice versa. Since reciprocal slopes indicate perpendicular lines, the lines 1 and 2 are perpendicular to each other.

Hope this helps!

5 0

3 years ago

Barry Road his scooter<br> for 24 minutes at a speed of 0.5 per minute how far did he go

liraira [26]

Well, I'm not sure what units "0.5" apply to, but to find the distance he rode, you will need to multiply.

0.5 * 24 = 12.

Barry rode his scooter for a distance of 12 [insert units here].

Hope this helps! Have a great day :)

7 0

2 years ago

Plzzzzzz help right answer gets brainly

GuDViN [60]

Answer:

3 * 3 * 3 * 3

=> $3^{4}$

If my answer helped, then kindly mark me as the brainliest!!

Thank You!!

6 0

3 years ago