How do you find an equation of the tangent line to the curve at the given point $y = sec x - 2 cos x$ $y=secx - 2cosx$ and $(\frac{π}{3}, 1)$ $(pi/3, 1)$ ?

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How do you find an equation of the tangent line to the curve at the given point $y = sec x - 2 cos x$ $y=secx - 2cosx$ and $(\frac{π}{3}, 1)$ $(pi/3, 1)$ ?

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Answer 1 · 2018-04-25T17:25:31

$y = 3 \sqrt{3} x - π \sqrt{3} + 1$ $y=3sqrt3x-pisqrt3+1$

Explanation:

$the slope of the tangent m = \frac{d y}{d x} at x = \frac{π}{3}$ $"the slope of the tangent m "=dy/dx" at "x=pi/3$

$\Rightarrow \frac{d y}{d x} = sec x tan x + 2 sin x$ $rArrdy/dx=secxtanx+2sinx$

$\frac{d y}{d x} (x = \frac{π}{3}) = sec (\frac{π}{3}) tan (\frac{π}{3}) + 2 sin (\frac{π}{3})$ $dy/dx(x=pi/3)=sec(pi/3)tan(pi/3)+2sin(pi/3)$

$\times \times \times \times x = 2 \times \sqrt{3} + 2 \times \frac{\sqrt{3}}{2}$ $color(white)(xxxxxxxxx)=2xxsqrt3+2xxsqrt3/2$

$\times \times \times \times x = 2 \sqrt{3} + \sqrt{3} = 3 \sqrt{3}$ $color(white)(xxxxxxxxx)=2sqrt3+sqrt3=3sqrt3$

$\Rightarrow y - 1 = 3 \sqrt{3} (x - \frac{π}{3})$ $rArry-1=3sqrt3(x-pi/3)$

$\Rightarrow y = 3 \sqrt{3} x - π \sqrt{3} + 1 \leftarrow equation of tangent$ $rArry=3sqrt3x-pisqrt3+1larrcolor(red)"equation of tangent"$

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How do you find an equation of the tangent line to the curve at the given point $y = sec x - 2 cos x$ $y=secx - 2cosx$ and $(\frac{π}{3}, 1)$ $(pi/3, 1)$ ?

1 Answer

Explanation:

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How do you find an equation of the tangent line to the curve at the given point y=secx−2cosxy=secx - 2cosx and (π3,1)(pi/3, 1)?

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How do you find an equation of the tangent line to the curve at the given point $y = sec x - 2 cos x$ $y=secx - 2cosx$ and $(\frac{π}{3}, 1)$ $(pi/3, 1)$ ?