# Find the equation of the tangent line to the curve y=5sec(x)-10cos(x) at the point (pi/3,5).The equation of this tangent line can be written in the form y=mx+b. Find m and b

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You need to remember how the equation of tangent line to a curve at a point looks like such that:

`y - y_0 = f'(x_0)(x - x_0)`

You need to identify what are the `x_0` and `y_0` coordinates of the point of tangency, hence, the problem provides that `x_0=pi/3` and `y_0=5` .

You need to differentiate the function with respect to x such that:

`f'(x) = (5/(cos x) - 10 cos x)'`

`f'(x) = 5sin x/(cos^2 x) + 10 sin x`

You need to substitute`pi/3` for x in f'(x) such that:

`f'(pi/3) = 5(sin (pi/3))/(cos^2 (pi/3)) + 10 sin (pi/3)`

Since `sin (pi/3) = sqrt3/2` and `cos(pi/3) = 1/2` yields:

`f'(pi/3) = 5(sqrt3/2)/(1/4) + 10 sqrt3/2`

`f'(pi/3) = 10 sqrt3 + 10 sqrt3/2`

`f'(pi/3) =30 sqrt3/2 =gt f'(pi/3) = 15sqrt3`

You need to substitute `pi/3` for `x_0` , 5 for `y_0` and `15sqrt3` for `f'(x_0) ` in equation of tangent line such that:

`y - 5 = 15sqrt3(x - pi/3)`

You need to use the slope intercept form of equation of tangent line, hence, you need to isolate y to the left side such that:

`y = 15sqrt3x - 5sqrt3 + 5`

**Hence, evaluating the equation of tangent line to the graph `y = 5/(cos x) - 10 cos x` , at `(pi/3,5)` yields `y = 15sqrt3x - 5(sqrt3 - 1)` .**