Animated Solution for Mathematics - Inverse Trigonometric Functions: 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

The given equation is:

To solve for $\alpha$, we must establish a common language by converting all terms into ${\tan }^{-1}$. Consider the term ${\cos }^{-1}\left(\frac{4}{5}\right)$. If we let $\theta ={\cos }^{-1}\left(\frac{4}{5}\right)$, then $\cos \theta =\frac{4}{5}$.

Using a right-angled triangle with base $4$ and hypotenuse $5$, the perpendicular side is $\sqrt{5^2-4^2}=3$. Therefore, $\tan \theta =\frac{3}{4}$, which implies $\theta ={\tan }^{-1}\left(\frac{3}{4}\right)$.

Substituting this back into our original equation, we get:

Isolating the ${\sin }^{-1}$ term, we have:

Applying the identity ${\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)$, we substitute $x=\frac{77}{36}$ and $y=\frac{3}{4}$:

Thus, ${\sin }^{-1}\left(\frac{\alpha }{17}\right)={\tan }^{-1}\left(\frac{8}{15}\right)$. Converting ${\tan }^{-1}\left(\frac{8}{15}\right)$ back to ${\sin }^{-1}$ using a triangle with perpendicular $8$ and base $15$ (hypotenuse $17$), we find ${\sin }^{-1}\left(\frac{8}{17}\right)$. Comparing the arguments, we conclude $\alpha =8$.

We now evaluate ${\sin }^{-1}(\sin 8)+{\cos }^{-1}(\cos 8)$. Since $8$ is in radians and $2\pi \approx 6.28$ while $3\pi \approx 9.42$, the value $8$ lies in the interval $(2\pi ,3\pi )$.

For ${\sin }^{-1}(\sin x)$, the function behaves as $y=3\pi -x$ in the interval $[2.5\pi ,3.5\pi ]$. Since $8\approx 8$, which is less than $2.5\pi \approx 7.85$, we must be careful; however, checking the range, $8$ falls into the branch where ${\sin }^{-1}(\sin 8)=3\pi -8$.

For ${\cos }^{-1}(\cos x)$, the function behaves as $y=x-2\pi$ in the interval $[2\pi ,3\pi ]$. Thus, ${\cos }^{-1}(\cos 8)=8-2\pi$.

Summing these results:

The final answer is $\pi$.