1.
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\caption{Figure 1}
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The point \(A\) is a distance 1 unit from the fixed origin \(O\) ．Its position vector is \(\mathbf { a } = \frac { 1 } { \sqrt { 2 } } ( \mathbf { i } + \mathbf { j } )\) ． The point \(B\) has position vector \(\mathbf { a } + \mathbf { j }\) ，as shown in Figure 1.
By considering \(\triangle O A B\) ，prove that \(\tan \frac { 3 \pi } { 8 } = 1 + \sqrt { } 2\) ．