If $\sin ^{-1} \frac{\alpha}{17}+\cos ^{-1} \frac{4}{5}-\tan ^{-1} \frac{77}{36}=0,0<\alpha<13$, then $\sin ^{-1}(\sin \alpha)+\cos ^{-1}(\cos \alpha)$ is equal to
Step 1: Start with the given equation: \[ \cos^{-1} \left( \frac{4}{5} \right) = \tan^{-1} \left( \frac{3}{4} \right). \] This equation involves the inverse trigonometric functions, which relate the angles whose cosine and tangent are given. We will use trigonometric identities and inverse function properties to simplify and solve the problem.
Step 2: Now, express \( \sin^{-1} \left( \frac{\alpha}{17} \right) \): \[ \sin^{-1} \left( \frac{\alpha}{17} \right) = \tan^{-1} \left( \frac{77}{36} \right) - \tan^{-1} \left( \frac{3}{4} \right). \] Here, we need to find the value of \( \alpha \) in terms of known constants by using the difference of tangents formula.
Step 3: Using the identity for the difference of tangents: \[ \tan^{-1} \left( \frac{A}{B} \right) - \tan^{-1} \left( \frac{C}{D} \right) = \tan^{-1} \left( \frac{AD - BC}{BD} \right), \] we apply it to the given equation to simplify: \[ \sin^{-1} \left( \frac{\alpha}{17} \right) = \tan^{-1} \left( \frac{77 \times 4 - 36 \times 3}{36 \times 4} \right) = \tan^{-1} \left( \frac{8}{15} \right). \] The identity helps us express the difference of the two inverse tangents as a single term, simplifying the equation.
Step 4: Hence, we obtain: \[ \sin^{-1} \left( \frac{\alpha}{17} \right) = \sin^{-1} \left( \frac{8}{17} \right). \] From this, we deduce that \( \alpha = 8 \), as the arcsine of \( \frac{\alpha}{17} \) equals the arcsine of \( \frac{8}{17} \).
Step 5: Therefore, we conclude: \[ \sin^{-1} (\sin 8) + \cos^{-1} (\cos 8) = 3\pi - 8 + 8 - 2\pi = \pi. \] This final step simplifies the trigonometric expressions and shows that the sum of the inverse sine and inverse cosine of the same angle is equal to \( \pi \), confirming the result.