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(B) Given: $x + \sqrt{x^2 + 1} = a$  Let $x + \sqrt{x^2 + 1} = a$  --- $(1)$  We know that $(\sqrt{x^2 + 1} + x)(\sqrt{x^2 + 1} - x) = (x^2 + 1) - x^2

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$\frac{1}{2}(a - \frac{1}{a})$

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(B) Given: $x + \sqrt{x^2 + 1} = a$  Let $x + \sqrt{x^2 + 1} = a$  --- $(1)$  We know that $(\sqrt{x^2 + 1} + x)(\sqrt{x^2 + 1} - x) = (x^2 + 1) - x^2 = 1$  So,$a(\sqrt{x^2 + 1} - x) = 1$  $\sqrt{x^2 + 1} - x = \frac{1}{a}$  --- $(2)$  Subtracting equation $(2)$ from equation $(1)$:  $(x + \sqrt{x^2 + 1}) - (\sqrt{x^2 + 1} - x) = a - \frac{1}{a}$  $2x = a - \frac{1}{a}$  $x = \frac{1}{2}(a - \frac{1}{a})$

If $x + \sqrt{x^2 + 1} = a$,then $x =$

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