# Show that the eq.n of the line passing through (acos^3 θ, asin^3 θ) and to the line xsecθ+ycosecθ = a is xcosθ - ysinθ = acos2θ

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The problem provies the information that the line passes through the point (`acos^3 theta, a sin^3 theta` ) and it intersects the line `x sec theta + y csc theta = a` .

Supposing that the line is `xcos theta - ysin theta = a cos 2 theta` , then, you should check if the coordinates of the point (`acos^3 theta, a sin^3 theta` ) verify the equation `xcos theta - ysin theta = acos 2 theta` such that:

`acos^3 theta*cos theta - a sin^3 theta*sin theta = a cos 2 theta`

You may divide by a both sides such that:

`cos^4 theta - sin^4 theta = cos 2 theta`

You need to write `cos^4 theta - sin^4 theta` as a difference of two squares such that:

`(cos^2 theta - sin^2 theta)(cos^2 theta+ sin^2 theta) = cos 2 theta`

Using the fundamental identity of trigonometry `cos^2 theta + sin^2 theta = 1` yields:

`cos^2 theta - sin^2 theta = cos 2 theta`

`cos 2 theta = cos 2 theta`

You need to put the equation `x sec theta + y csc theta = a` in the slope intercept form such that:

`x sec theta + y csc theta = a`

`y* csc theta = -x sec theta + a`

`y = -x*(1/cos theta)/(1/sin theta) + a`

`y = -x*tan theta + a`

You need to identify the slope of the equation as `m_1 = -tan theta` .

You need to put the equation `xcos theta - ysin theta = acos 2 theta` in the slope intercept form such that:

`ysin theta = x cos theta + a cos 2 theta`

`y = x*cos theta/sin theta + a cos 2 theta/sin theta`

`y = x*cot theta + a cos 2 theta/sin theta`

You need to identify the slope of the equation as `m_2 =cot theta` .

Notice that multiplying the slopes yields:

`m_1*m_2 = -tan theta*cot theta`

You should remember that `cot theta = 1/tan theta` such that:

`m_1*m_2 = -tan theta*(1/tan theta) `

`m_1*m_2 =-1`

Since multiplying the slopes yields -1, it means that the lines are perpendicular, hence the line `xcos theta - ysin theta = acos 2 theta ` passes through the line `x sec theta + y csc theta = a` .

**Hence, substituting the coordinates of the point `(acos^3 theta, a sin^3 theta)` in the equation`xcos theta - ysin theta = acos 2 theta ` , they verify the equation and the line `x sec theta + y csc theta = a` is perpendicular to`xcos theta - ysin theta = acos 2 theta` .**

**Sources:**

Let,

EQ.1: xsecθ + ycscθ = a

EQ.2: xcosθ - ysinθ = acos2

> Determine the slope of each equation.

For EQ.1: xsecθ + ycscθ = a

ycscθ = -xsecθ + a

(ycscθ = -xsecθ + a) sinθ

y = -xsecθsinθ + asinθ

y = -xtanθ + asinθ

So slope, m1 = -tanθ.

For EQ. 2: xcosθ - ysinθ = acos2

- ysinθ = -xcosθ + acos2θ

( - ysinθ = -xcosθ + acos2θ ) * (1/-sinθ)

y = xcosθ/sinθ + acos2θ/sinθ

y = xcotθ + acos2θ/sinθ

So slope, m2 = cotθ

*Since, the slopes m1 and m2 are negative reciprocal to each other, it means EQ1 is perpendicular to EQ.2. Hence, the two lines inersect. *

*>*Determine if EQ.2 contains the point (acos^3 θ, asin^3 θ).* *

Substitute (acos^3 θ, asin^3 θ) to EQ.2, and determine if it will result to a True condition.

xcosθ - ysinθ = acos2θ

(acos^3 θ)cosθ - (asin^3 θ)sinθ = acos2θ

acos^4 θ - asin^4 θ = acos2θ

a(cos^4 θ - sin^4 θ) = acos2θ

a(cos^2 θ + sin^2 θ) ( cos^2 θ - sin^2θ) = acos2θ

Then, simplify left side by using the pythagorean and double angle identities of trigonometry.

* *a (1) (cos2θ) = acos2θ

acos2θ = acos2θ (TRUE)

Hence, EQ.2 contains the point (acos^3 θ, asin^3 θ).

Answer: xcosθ - ysinθ = acos2 is the line that passes through (acos^3 **θ, **asin^3 θ) and to the line xsec**θ+ycosec θ = a .**