# How do you solve #cot [Arcsin (-12/13)]#?

value of a is in the 4th quadrant. The general value is either in the

The given expression is cot a = cos a/sin a and, accordingly, the

answer is given..

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To solve ( \cot(\arcsin(-\frac{12}{13})) ), first, find the value of ( \arcsin(-\frac{12}{13}) ), then take the cotangent of that value.

Given ( \arcsin(-\frac{12}{13}) ), we need to find the angle whose sine is ( -\frac{12}{13} ). Since sine is negative in the third and fourth quadrants, we're looking for an angle in either the third or fourth quadrant.

Let's call the angle ( \theta ). We know that ( \sin(\theta) = -\frac{12}{13} ). Using the Pythagorean identity ( \sin^2(\theta) + \cos^2(\theta) = 1 ), we can find ( \cos(\theta) ).

[ \cos^2(\theta) = 1 - \sin^2(\theta) = 1 - \left(-\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169} ] [ \cos(\theta) = \pm \sqrt{\frac{25}{169}} = \pm \frac{5}{13} ]

Since cosine is also negative in the third quadrant, ( \cos(\theta) = -\frac{5}{13} ).

Now, to find ( \cot(\arcsin(-\frac{12}{13})) ), we use the fact that ( \cot(\theta) = \frac{1}{\tan(\theta)} ), and ( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} ).

[ \tan(\theta) = \frac{-\frac{12}{13}}{-\frac{5}{13}} = \frac{12}{5} ]

Thus, ( \cot(\arcsin(-\frac{12}{13})) = \frac{1}{\tan(\arcsin(-\frac{12}{13}))} = \frac{1}{\frac{12}{5}} = \frac{5}{12} ).

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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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