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mote1985 [20]
3 years ago
10

Sec^6x(secxtanx)-sec^4(secxtanx)=sec^5xtan^3x Verify the trigonometric identity

Mathematics
1 answer:
mario62 [17]3 years ago
7 0

I guess you mean

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

On the left side, we have a common factor of \sec^4x(\sec x\tan x)=\sec^5x\tan x, so that

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

Recall that

\sec^2x=1+\tan^2x

from which it follows that

\sec^5x\tan x(\sec^2x-1)=\sec^5x\tan x\tan^2x=\sec^5x\tan^3x

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4 years ago
What is an equation of the line that passes through the points (-6, -2) and
den301095 [7]

Answer:

The equation of line is: \mathbf{4x-3y=-18}

Step-by-step explanation:

We need to find an equation of the line that passes through the points (-6, -2) and  (-3, 2)?

The equation of line in slope-intercept form is: y=mx+b

where m is slope and b is y-intercept.

We need to find slope and y-intercept.

Finding Slope

Slope can be found using formula: Slope=\frac{y_2-y_1}{x_2-x_1}

We have x_1=-6,y_1=-2, x_2=-3, y_2=2

Putting values and finding slope

Slope=\frac{2-(-2)}{-3-(-6)}\\Slope=\frac{2+2}{-3+6} \\Slope=\frac{4}{3}

So, we get slope: m=\frac{4}{3}

Finding y-intercept

Using point (-6,-2) and slope m=\frac{4}{3} we can find y-intercept

y=mx+b\\-2=\frac{4}{3}(-6)+b\\-2=4(-2)+b\\-2=-8+b\\b=-2+8\\b=6

So, we get y-intercept b= 6

Equation of required line

The equation of required line having slope m=\frac{4}{3} and y-intercept b = 6 is

y=mx+b\\y=\frac{4}{3}x+6

Now transforming in fully reduced form:

y=\frac{4x+6*3}{3} \\y=\frac{4x+18}{3} \\3y=4x+18\\4x-3y=-18

So, the equation of line is: \mathbf{4x-3y=-18}

6 0
3 years ago
A trail has markers every 1/8 mile. Jody starts at the 2 1/4 mile marker, hikes to the 4 3/8 marker, and then hikes back to the
vlada-n [284]
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At the library, 56 people borrowed
irina [24]

Answer:

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Every day, the number of books is being subtracted by 2.

56 - Monday (-8)

48 - Tuesday (-8)

40 - Wednesday (-8)

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5 0
3 years ago
A rectangular box has a square base. The combined length of a side of the square base, and the height is 20 in. Let x be the len
aniked [119]

Answer:

a. V = (20-x) x^{2} in^{3}  

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b. What is the maximum possible volume of the box?

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Let dV / Dx = 0, we have:

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<=> x = 40/3

=>the height h = 20/3

So  the maximum possible volume of the box is:

V = 20/3 * 40/3 *40/3

= 1185.185 in^{3}

7 0
3 years ago
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