Write tan(2-i) in the form u+iv
Write $\tan(2-i)$ in the form $u+iv$.
Explanation of Method
Use the identity $\tan(x+iy)=\dfrac{\sin(2x)+i\sinh(2y)}{\cos(2x)+\cosh(2y)}$, obtained from $\tan z=\sin z/\cos z$ together with the double-angle expansion of $\sin z\overline{\cos z}$-type products, then substitute and simplify.
Step 1 · The identity for complex tangent
$$ \tan(x+iy)=\frac{\sin(2x)+i\sinh(2y)}{\cos(2x)+\cosh(2y)} $$
Step 2 · Apply with $x=2,\ y=-1$
$$ \tan(2-i)=\frac{\sin(4)+i\sinh(-2)}{\cos(4)+\cosh(-2)}=\frac{\sin(4)-i\sinh(2)}{\cos(4)+\cosh(2)} $$
(using $\sinh(-2)=-\sinh(2)$ and $\cosh(-2)=\cosh(2)$.)
Step 3 · Numerical evaluation
$$ \sin(4)\approx-0.756802,\quad \cos(4)\approx-0.653644,\quad \sinh(2)\approx3.626860,\quad \cosh(2)\approx3.762196 $$
Denominator: $\cos(4)+\cosh(2)\approx3.108552$ (real and positive, since $\cosh2>|\cos4|$).
$$ u=\frac{\sin4}{\cos4+\cosh2}\approx-0.243458, \qquad v=\frac{-\sinh2}{\cos4+\cosh2}\approx-1.166736 $$
Step 4 · Result
$$ \boxed{\tan(2-i)=\frac{\sin4-i\sinh2}{\cos4+\cosh2}\approx-0.2435-1.1667i} $$
Verification
Direct numerical evaluation of $\tan(2-i)=\sin(2-i)/\cos(2-i)$ using the complex sine and cosine definitions reproduces $-0.243458\ldots-1.166736\ldots i$ to full precision, confirming the identity.
Result
The exact real and imaginary parts are $u=\dfrac{\sin4}{\cos4+\cosh2}$ and $v=\dfrac{-\sinh2}{\cos4+\cosh2}$.