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4 years ago

5

Tan x/sec^2 x + tan x/csc^2 x

1 answer:

Arada [10]4 years ago

5 0

I guess you're supposed to simplify?

$\dfrac{\tan x}{\sec^2x}+\dfrac{\tan x}{\csc^2x}$

$=\tan x\left(\dfrac1{\sec^2x}+\dfrac1{\csc^2x}\right)$

$=\tan x(\cos^2x+\sin^2x)$

$=\tan x$

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the line containing the altitude to the hypotenuse of a right triangle whose vertices are p(-1, 1), q(3, 5), and r(5, -5).

finlep [7]

The equation of the straight line is given by 5y-x=6

An equation can be used to describe a straight line drawn on the Cartesian Plane. These equations have a generic structure and can change based on the slope and where the line intersects the axes.

We will greatly simplify this issue by utilizing the fact that this triangle is specified as a right triangle. We would have had to establish the presence of a right angle if it had been presented as a general triangle.

The segment QR is the hypotenuse of the triangle.

Any side's altitude is located along a line that is perpendicular to that side. We are aware that the slopes of perpendicular lines are negative reciprocals, or:

Slope of line joining Q(3, 5) and R(5, -5)

m= $\frac{y_2-y_1}{x_2-x_1}$

Hence slope is= -10/2= -5

Now we know that for two perpendicular line the slope of one line is the negative reciprocal of the other.

Slope of line perpendicular to QR=1/5

This line passes through P(-1,1)

Equation of the straight line that passes through (1,-1) and with a slope of 1/5 is given by:

$(y-y_1)=m(x-x_1)\\$

Substituting the values we get:

y-1=1/5(x+1)

or,5y-5=x+1

or,5y-x=6

Hence the equation of the required line is : 5y-x=6

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3 years ago

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Correct answer is A.

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Answer:(6,-2)

Step-by-step explanation:

M is the midpoint of AB then ,

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so, B= (6,-2)

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