Find the value of sin-1⁡(513)+cos-1⁡(35)

The correct option is sin⁻¹(63/65).

To find the value of the expression ${\sin }^{-1}\left(\frac{5}{13}\right)+{\cos }^{-1}\left(\frac{3}{5}\right)$, we can express both inverse trigonometric terms in terms of sine or cosine to apply a standard addition formula. Let us convert ${\cos }^{-1}\left(\frac{3}{5}\right)$ into a sine inverse term.

Let $A={\sin }^{-1}\left(\frac{5}{13}\right)$ and $B={\cos }^{-1}\left(\frac{3}{5}\right)$.
From these definitions, we have:
$\sin \left(A\right)=\frac{5}{13}$
$\cos \left(B\right)=\frac{3}{5}$

Since $A$ and $B$ correspond to acute angles in right-angled triangles, we can find the corresponding cosine and sine values using the Pythagorean identity ${\sin }^2\left(\theta \right)+{\cos }^2\left(\theta \right)=1$:
$\cos \left(A\right)=\sqrt{1-{\sin }^2\left(A\right)}=\sqrt{1-{\left(\frac{5}{13}\right)}^2}=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}$
$\sin \left(B\right)=\sqrt{1-{\cos }^2\left(B\right)}=\sqrt{1-{\left(\frac{3}{5}\right)}^2}=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$

Now we want to find the value of $A+B$. We can calculate $\sin (A+B)$ using the sine addition formula:
$\sin (A+B)=\sin \left(A\right)\cos \left(B\right)+\cos \left(A\right)\sin \left(B\right)$

Substitute the values we calculated:
$\sin (A+B)=\left(\frac{5}{13}\right)·\left(\frac{3}{5}\right)+\left(\frac{12}{13}\right)·\left(\frac{4}{5}\right)$
$\sin (A+B)=\frac{15}{65}+\frac{48}{65}$
$\sin (A+B)=\frac{63}{65}$

Taking the inverse sine of both sides, we get:
$A+B={\sin }^{-1}\left(\frac{63}{65}\right)$

Therefore, the value of the given expression is:
${\sin }^{-1}\left(\frac{5}{13}\right)+{\cos }^{-1}\left(\frac{3}{5}\right)={\sin }^{-1}\left(\frac{63}{65}\right)$