If $$\sin^{-1}\dfrac{\alpha}{17} + \cos^{-1}\dfrac{4}{5} - \tan^{-1}\dfrac{77}{36} = 0$$, $$0 < \alpha < 13$$, then $$\sin^{-1}\sin\alpha + \cos^{-1}\cos\alpha$$ is equal to

Given $$\sin^{-1}\frac{\alpha}{17} + \cos^{-1}\frac{4}{5} - \tan^{-1}\frac{77}{36} = 0$$ with $$0 < \alpha < 13$$, we need to find $$\sin^{-1}(\sin\alpha) + \cos^{-1}(\cos\alpha)$$.

To begin,

Rearranging: $$\sin^{-1}\frac{\alpha}{17} = \tan^{-1}\frac{77}{36} - \cos^{-1}\frac{4}{5}$$

Convert $$\cos^{-1}\frac{4}{5}$$ to inverse tangent. If $$\cos\theta = 4/5$$, then $$\sin\theta = 3/5$$ (for $$\theta \in [0, \pi/2]$$), so:

$$ \cos^{-1}\frac{4}{5} = \tan^{-1}\frac{3}{4} $$

Now use the subtraction formula $$\tan^{-1}a - \tan^{-1}b = \tan^{-1}\frac{a-b}{1+ab}$$ (valid when $$ab > -1$$):

$$ \tan^{-1}\frac{77}{36} - \tan^{-1}\frac{3}{4} = \tan^{-1}\left(\frac{\frac{77}{36} - \frac{3}{4}}{1 + \frac{77}{36} \cdot \frac{3}{4}}\right) $$

Numerator: $$\frac{77}{36} - \frac{3}{4} = \frac{77 - 27}{36} = \frac{50}{36} = \frac{25}{18}$$

Denominator: $$1 + \frac{77 \times 3}{36 \times 4} = 1 + \frac{231}{144} = \frac{144 + 231}{144} = \frac{375}{144} = \frac{125}{48}$$

$$ = \tan^{-1}\left(\frac{25/18}{125/48}\right) = \tan^{-1}\left(\frac{25}{18} \times \frac{48}{125}\right) = \tan^{-1}\left(\frac{1200}{2250}\right) = \tan^{-1}\frac{8}{15} $$

So $$\sin^{-1}\frac{\alpha}{17} = \tan^{-1}\frac{8}{15}$$. If $$\tan\theta = 8/15$$, then $$\sin\theta = \frac{8}{\sqrt{64+225}} = \frac{8}{\sqrt{289}} = \frac{8}{17}$$.

$$ \sin^{-1}\frac{\alpha}{17} = \sin^{-1}\frac{8}{17} \implies \alpha = 8 $$

Next,

Since $$\alpha = 8$$ radians, we need to use the principal value properties.

For $$\sin^{-1}(\sin 8)$$: Since $$8 \approx 2\pi + 1.717$$ and $$8 - 2\pi \approx 1.717$$, which lies in $$(\pi/2, \pi)$$. So $$\sin^{-1}(\sin 8) = \sin^{-1}(\sin(8 - 2\pi)) = \pi - (8 - 2\pi) = 3\pi - 8$$.

For $$\cos^{-1}(\cos 8)$$: Since $$8 \approx 2\pi + 1.717$$, and $$\cos(8) = \cos(8 - 2\pi)$$. Since $$8 - 2\pi \approx 1.717 \in (0, \pi)$$, we have $$\cos^{-1}(\cos 8) = 8 - 2\pi$$.

From this,

$$ \sin^{-1}(\sin 8) + \cos^{-1}(\cos 8) = (3\pi - 8) + (8 - 2\pi) = \pi $$

The answer is $$\pi$$.

The correct answer is Option 1: $$\pi$$.