# How do you simplify the expression #sec(arccos (-3/5))#?

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Use cos arc cos x = x.

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To simplify the expression ( \sec(\arccos(-\frac{3}{5})) ), you can use the identity relating secant and cosine functions.

Since ( \cos(\theta) = \frac{adjacent}{hypotenuse} ), and ( \arccos(\frac{-3}{5}) ) yields an angle whose cosine is ( -\frac{3}{5} ), you can use a right triangle to represent this situation.

Let ( \theta = \arccos(-\frac{3}{5}) ), then ( \cos(\theta) = -\frac{3}{5} ). Consider a right triangle where the adjacent side is ( -3 ) and the hypotenuse is ( 5 ). Using the Pythagorean theorem, you can find the length of the opposite side to be ( 4 ).

Now, ( \sec(\theta) = \frac{hypotenuse}{adjacent} ), so ( \sec(\arccos(-\frac{3}{5})) = \frac{5}{-3} = -\frac{5}{3} ). Therefore, the simplified expression is ( -\frac{5}{3} ).

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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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