Concept:
One of the most useful inverse trigonometric identities is
\[
\sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2},
\qquad -1\le x\le1.
\]
This identity allows us to simplify expressions involving inverse sine and inverse cosine functions having the same argument.
Step 1: Apply the inverse trigonometric identity.
Using
\[
\sin^{-1}(x)+\cos^{-1}(x)=\frac{\pi}{2},
\]
with
\[
x=-\frac12,
\]
we obtain
\[
\cos^{-1}\left(-\frac12\right)
+
\sin^{-1}\left(-\frac12\right)
=
\frac{\pi}{2}.
\]
Step 2: Evaluate \(\tan^{-1}(1)\).
We know that
\[
\tan\left(\frac{\pi}{4}\right)=1.
\]
Therefore,
\[
\tan^{-1}(1)=\frac{\pi}{4}.
\]
Step 3: Add the obtained values.
\[
\frac{\pi}{4}+\frac{\pi}{2}
=
\frac{\pi}{4}+\frac{2\pi}{4}
=
\frac{3\pi}{4}.
\]
Step 4: Verification.
\[
\cos^{-1}\left(-\frac12\right)=\frac{2\pi}{3},
\qquad
\sin^{-1}\left(-\frac12\right)=-\frac{\pi}{6}.
\]
Hence,
\[
\frac{2\pi}{3}-\frac{\pi}{6}
=
\frac{\pi}{2},
\]
which confirms the result.
Conclusion:
Therefore,
\[
\tan^{-1}(1)
+
\cos^{-1}\left(-\frac12\right)
+
\sin^{-1}\left(-\frac12\right)
=
\boxed{\frac{3\pi}{4}}.
\]
Hence, the correct option is
\[
\boxed{(C)}.
\]