What is the equation of the tangent line of $r = {tan}^{2} (θ) - sin (θ - π)$ $r=tan^2(theta) - sin(theta-pi)$ at $θ = \frac{π}{4}$ $theta=pi/4$ ?

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Answer 1 · 2018-05-10T00:01:09

$r = \frac{2 + \sqrt{2}}{2}$ $r=(2+sqrt2)/2$

$r = {tan}^{2} θ - sin (θ - π)$ $r=tan^2 theta- sin(theta - pi)$ at $\frac{π}{4}$ $pi/4$

$r = {tan}^{2} (\frac{π}{4}) - sin (\frac{π}{4} - π)$ $r=tan^2 (pi/4) - sin (pi/4 -pi)$

$r = 1^{2} - sin (\frac{- 3 π}{4})$ $r=1^2 - sin((-3pi)/4)$

$r = 1 - sin (\frac{5 π}{4})$ $r=1-sin((5pi)/4)$

$r = 1 - (- \frac{\sqrt{2}}{2})$ $r=1-(-sqrt2/2)$

$r = 1 + \frac{\sqrt{2}}{2}$ $r=1+sqrt2/2$

$r = \frac{2 + \sqrt{2}}{2}$ $r=(2+sqrt2)/2$

What is the equation of the tangent line of r=tan2(θ)−sin(θ−π)r=tan^2(theta) - sin(theta-pi) at θ=π4theta=pi/4?