Answer
\begin{array}{l}
(a)=1 \\
(b)=\frac{n(1+n)(1+2n)}{6}
\end{array}
Work Step by Step
(a) We are given that
\[
f(u, v, \lambda, \phi)=e^{u+v} \tan \lambda \cos \phi
\]
We have to find $f\left(-2,2,0, \frac{\pi}{4}\right),$ substituting $u, v, \lambda$ and $\phi$ in the function
\[
\begin{aligned}
f(u, v, \lambda, \phi)=e^{u+v} \tan \phi \cos \lambda & \\
f\left(-2,2,0, \frac{\pi}{4}\right) &=e^{-2+2} \tan \left(\frac{\pi}{4}\right) \\ \cos (0)
f\left(-2,2,0, \frac{\pi}{4}\right) &=1 * 1 * 1 \\
f\left(-2,2,0, \frac{\pi}{4}\right) &=1
\end{aligned}
\]
(b) We are given that
\[
f\left(x_{1}, x_{2}, x_{3}, \ldots x_{n}\right)=x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}+\ldots . .+x_{n}^{2}
\]
We have to find $f(1,2,3, \dots, n)$
\[
\begin{array}{l}
f(1,2,3 \ldots . n)=1^{2}+2^{2}+3^{2}+4^{2} \ldots . \\
f(1,2,3 \ldots . n)=\sum_{n=1}^{n} n^{2} \\
f(1,2,3 \ldots . n)=\frac{n(n+1)(2 n+1)}{6}
\end{array}
\]
