Q.Show that sin⁻¹(4/5) + sin⁻¹(5/13) + sin⁻¹(16/65) = π/2.

Combine the first two inverse-sine terms into $\tan^{-1}(63/16)$, then show this equals $\pi/2$ minus the third term.

Let $A=\sin^{-1}\tfrac45$ so $\sin A=\tfrac45,\cos A=\tfrac35,\tan A=\tfrac43$ (using $\cos A=\sqrt{1-\sin^2A}$, both positive since $A$ is acute).

Let $B=\sin^{-1}\tfrac{5}{13}$ so $\sin B=\tfrac5{13},\cos B=\tfrac{12}{13},\tan B=\tfrac{5}{12}$.

Combine using the tangent addition formula:

$\tan(A+B) = \frac{\tan A+\tan B}{1-\tan A\tan B} = \frac{\tfrac43+\tfrac5{12}}{1-\tfrac43\cdot\tfrac5{12}} = \frac{\tfrac{21}{12}}{\tfrac{16}{36}} = \frac{63}{16}$

Let $C=\sin^{-1}\tfrac{16}{65}$, so $\sin C=\tfrac{16}{65},\ \cos C=\sqrt{1-\tfrac{256}{4225}}=\tfrac{63}{65}$ (using $63^2+16^2=3969+256=4225=65^2$).

Then $\cot C = \dfrac{\cos C}{\sin C} = \dfrac{63/65}{16/65}=\dfrac{63}{16}$, i.e. $\tan\!\left(\dfrac{\pi}{2}-C\right)=\dfrac{63}{16}$.

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