Find the exact value of the expression.

[tex]\[ \tan \left(\sin ^{-1} \frac{12}{13}+\cos ^{-1}(-1)\right) \][/tex]

[tex]\[ \tan \left(\sin ^{-1} \frac{12}{13}+\cos ^{-1}(-1)\right)= \][/tex]

(Type an exact answer in simplified form.)



Answer :

To find the exact value of the expression [tex]\(\tan \left(\sin^{-1} \frac{12}{13} + \cos^{-1}(-1)\right)\)[/tex], follow these steps:

1. Understand the given values:
[tex]\[ \sin^{-1} \frac{12}{13} \quad \text{and} \quad \cos^{-1}(-1) \][/tex]

2. Evaluate [tex]\(\cos^{-1}(-1)\)[/tex]:
By definition, [tex]\(\cos^{-1}(-1)\)[/tex] is the angle whose cosine is [tex]\(-1\)[/tex]. The angle whose cosine is [tex]\(-1\)[/tex] in the principal range [tex]\([0, \pi]\)[/tex] is [tex]\(\pi\)[/tex]:
[tex]\[ \cos^{-1}(-1) = \pi \][/tex]

3. Evaluate [tex]\(\sin^{-1} \frac{12}{13}\)[/tex]:
The angle whose sine value is [tex]\(\frac{12}{13}\)[/tex] in the principal range [tex]\([- \frac{\pi}{2}, \frac{\pi}{2}]\)[/tex]. Denote this angle as [tex]\(\theta\)[/tex]:
[tex]\[ \sin \theta = \frac{12}{13} \][/tex]

Using the Pythagorean identity [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex], we can determine [tex]\(\cos \theta\)[/tex]:
[tex]\[ \cos^2 \theta = 1 - \sin^2 \theta = 1 - \left(\frac{12}{13}\right)^2 = 1 - \frac{144}{169} = \frac{25}{169} \][/tex]
[tex]\[ \cos \theta = \sqrt{\frac{25}{169}} = \frac{5}{13} \][/tex]
Thus, [tex]\(\theta\)[/tex] satisfies:
[tex]\[ \theta = \sin^{-1} \frac{12}{13} \][/tex]

4. Sum the angles:
Next, we sum the angles that we have:
[tex]\[ \sin^{-1} \frac{12}{13} + \cos^{-1}(-1) = \theta + \pi \][/tex]

5. Evaluate [tex]\(\tan \left(\theta + \pi \right)\)[/tex]:
The tangent function has the property that [tex]\(\tan(x + \pi) = \tan x\)[/tex]:
[tex]\[ \tan(\theta + \pi) = \tan \theta \][/tex]

6. Determine [tex]\(\tan \theta\)[/tex]:
Recall that:
[tex]\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{5} \][/tex]

Thus, the exact value of the expression [tex]\(\tan \left(\sin^{-1} \frac{12}{13} + \cos^{-1}(-1)\right)\)[/tex] is:
[tex]\[ \boxed{\frac{12}{5}} \][/tex]