# How do you find the exact value of #sec(tan^-1(-3/5))#?

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To find the exact value of ( \sec(\tan^{-1}(-3/5)) ), we'll first use the definition of tangent inverse. Let ( \theta = \tan^{-1}(-3/5) ), which means ( \tan(\theta) = -\frac{3}{5} ).

Now, using the properties of right triangles, we can construct a right triangle with opposite side length -3 and adjacent side length 5. This gives us a hypotenuse of ( \sqrt{(-3)^2 + 5^2} = \sqrt{9 + 25} = \sqrt{34} ).

Now, we know that ( \sec(\theta) = \frac{1}{\cos(\theta)} ). Using our triangle, we can find ( \cos(\theta) ) by ( \frac{\text{adjacent}}{\text{hypotenuse}} ), so ( \cos(\theta) = \frac{5}{\sqrt{34}} ).

Therefore, ( \sec(\theta) = \frac{1}{\cos(\theta)} = \frac{\sqrt{34}}{5} ).

So, ( \sec(\tan^{-1}(-3/5)) = \frac{\sqrt{34}}{5} ).

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