# What is the equation of the tangent line of #r=tan^2theta - costheta# at #theta=pi/4#?

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To find the equation of the tangent line to the polar curve ( r = \tan^2(\theta) - \cos(\theta) ) at ( \theta = \frac{\pi}{4} ), first, we need to find the derivative of ( r ) with respect to ( \theta ), which gives us ( \frac{dr}{d\theta} ). Then, we evaluate ( r ) and ( \frac{dr}{d\theta} ) at ( \theta = \frac{\pi}{4} ) to get the slope of the tangent line. Finally, we use the point-slope form of a line to find the equation of the tangent line.

[ r = \tan^2(\theta) - \cos(\theta) ]

[ \frac{dr}{d\theta} = \frac{d}{d\theta}(\tan^2(\theta) - \cos(\theta)) ]

[ \frac{dr}{d\theta} = 2\tan(\theta)\sec^2(\theta) + \sin(\theta) ]

At ( \theta = \frac{\pi}{4} ):

[ r(\frac{\pi}{4}) = \tan^2(\frac{\pi}{4}) - \cos(\frac{\pi}{4}) = 1 - \frac{\sqrt{2}}{2} ]

[ \frac{dr}{d\theta}(\frac{\pi}{4}) = 2\tan(\frac{\pi}{4})\sec^2(\frac{\pi}{4}) + \sin(\frac{\pi}{4}) = 2\sqrt{2} + \frac{\sqrt{2}}{2} ]

So, the slope of the tangent line at ( \theta = \frac{\pi}{4} ) is ( m = 2\sqrt{2} + \frac{\sqrt{2}}{2} ).

Using the point-slope form of a line:

[ y - y_1 = m(x - x_1) ]

[ y - (1 - \frac{\sqrt{2}}{2}) = (2\sqrt{2} + \frac{\sqrt{2}}{2})(x - \frac{\pi}{4}) ]

[ y = (2\sqrt{2} + \frac{\sqrt{2}}{2})x + (1 - \frac{\sqrt{2}}{2}) - (2\sqrt{2} + \frac{\sqrt{2}}{2})\frac{\pi}{4} ]

[ y = (2\sqrt{2} + \frac{\sqrt{2}}{2})x + (1 - \frac{\sqrt{2}}{2}) - \frac{\pi}{2} ]

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