### Video Transcript

Find the equation of the tangent to
the curve π₯ equals five sec π and π¦ equals five tan π at π equals π by
six.

To find the equation of a tangent,
we must begin by calculating its gradient. That, of course, is the value of
the derivative at that point. So weβre going to need to begin by
calculating the value of dπ¦ by dπ₯ when π is equal to π by six.

Well, of course, these are
parametric equations. Thereβs an equation for π₯ in terms
of π and an equation for π¦ in terms of π. We recall that, given two
parametric equations β π₯ is some function in π and π¦ is some other function in π
β we can find dπ¦ by dπ₯ by dividing dπ¦ by dπ by dπ₯ by dπ. So we see weβre going to need to
begin by differentiating each of our functions with respect to π. Weβll use the general result for
the derivative of sec of π₯ and tan of π₯ to do so.

The derivative of sec of π₯ is sec
π₯ tan π₯. And the derivative of tan π₯ is sec
squared π₯. This means that dπ₯ by dπ is five
sec π tan π. And then dπ¦ by dπ is five sec
squared π. dπ¦ by dπ₯ is the quotient to these. Itβs five sec squared π divided by
five sec π tan π. And of course, we can simplify by
dividing through by five and by sec π. We then recall that sec π is the
same as one over cos π. We then recall that sec π is the
same as one over cos π. And weβre going to be dividing this
by tan π, or dividing by sin π over cos π.

Now dividing by a fraction is the
same as multiplying by the reciprocal of that fraction. So weβre going to be multiplying
one over cos π by cos π over sin π. And thatβs great because we have a
little more canceling to do. Weβve now fully simplified the
expression for the derivative of π¦ with respect π₯. Itβs one over sin π. We might write that as csc π.

Remember, we were looking to find
the value of the gradient of the tangent at π equals π by six. So letβs work out dπ¦ by dπ₯ when
π is π by six. Thatβs one over sin of π by
six. And since sin of π by six is
one-half, we find the gradient of the tangent to be two.

Now that we have the gradient to
the curve, letβs clear some space and see what else we need. The equation of a straight line
with the gradient of π that passes through a point with Cartesian coordinates π₯
one, π¦ one is π¦ minus π¦ one equals π times π₯ minus π₯ one. We quite clearly know the value of
π. But we donβt have an
π₯π¦-coordinate we can use.

But by substituting π equals π by
six into each of our original equations, weβll find the coordinate of the point
where the tangent meets the curve. We get π₯ equals five sec of π by
six. And actually thatβs 10 root three
over three. And then we get π¦ to be equal to
five tan of π by six. And thatβs five root three over
three.

Substituting what we have into our
formula for the equation of a straight line, and we get π¦ minus five root three
over three equals two times π₯ minus 10 root three over three. When we distribute the parentheses,
we get two π₯ minus 20 root three over three on the right-hand side. We rearrange by subtracting two π₯
and adding 20 root three over three to both sides to find the equation to be π¦
minus two π₯ plus 15 root three over three equals zero.

All thatβs left is for us to
simplify our fraction. And when we do, we find the
equation of the tangent to our curve when π is equal to π by six to be π¦ minus
two π₯ plus five root three equals zero.