# What is the slope of the line normal to the tangent line of #f(x) = cscx-sec^2x +1 # at # x= (17pi)/12 #?

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To find the slope of the line normal to the tangent line of ( f(x) = \csc(x) - \sec^2(x) + 1 ) at ( x = \frac{17\pi}{12} ), we first find the derivative of ( f(x) ) using the quotient rule and the chain rule. Then, we find the slope of the tangent line by evaluating the derivative at ( x = \frac{17\pi}{12} ). Finally, we find the slope of the normal line by taking the negative reciprocal of the slope of the tangent line.

The derivative of ( f(x) ) is:

[ f'(x) = -\csc(x) \cot(x) + 2\sec(x) \tan(x) ]

Evaluating ( f'(x) ) at ( x = \frac{17\pi}{12} ):

[ f'\left(\frac{17\pi}{12}\right) = -\csc\left(\frac{17\pi}{12}\right) \cot\left(\frac{17\pi}{12}\right) + 2\sec\left(\frac{17\pi}{12}\right) \tan\left(\frac{17\pi}{12}\right) ]

Then, find the slope of the normal line:

[ \text{Slope of normal line} = -\frac{1}{{f'\left(\frac{17\pi}{12}\right)}} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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