If a2+b2=1 then find the value of 1+b+ia/1+b-ia

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a^{2} + b^{2} = 1

\frac{1 + b + i a}{1 + b - i a} =

If $(a^{2} + b^{2}) = 1 \frac{1 + b + i a}{(1 + b - i a)}$

| \to A \times \to B | = \sqrt{3} \to A . \to B,

| \to A + \to B |

The value of $\frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} + \frac{1}{r_{3}^{2}} + \frac{1}{r^{2}}$ is

{(\frac{a}{b})}^{2} + {(\frac{b}{a})}^{2} - \frac{4}{i} (\frac{a}{b} - \frac{b}{a}) - 6

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