Question 131

If a (2+ √3) = b (2 - √3) then the value of $$1/(a^2+1)+1/(b^2+1)$$ is

Solution

Let $$a =(2-\sqrt{3})$$ and $$b =(2 + \sqrt{3})$$
$$a^{2}= 7- 4\sqrt{3}$$, and $$b^{2}= 7+4\sqrt{3}$$

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

$$(a^{2}+1)(b^{2}+1)$$ = $$(8-4\sqrt{3})(8+4\sqrt{3})$$ = $$16(2-\sqrt{3})(2+\sqrt{3}) = 16$$

$$a^{2} + b^{2} +2= 16$$
Thus $$ \frac{1}{a^{2}+1}+\frac{1}{b^{2}+1}=1$$

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