Question

If $	an A = (\sqrt{2} - 1)$, then prove that $\frac{	an A}{1 + 	an^2 A} = \frac{\sqrt{2}}{4}$.

Accepted Answer

Substitute $	an A = \sqrt{2} - 1$ into LHS: $$1 + 	an^2 A = 1 + (\sqrt{2} - 1)^2 = 1 + (2 + 1 - 2\sqrt{2}) = 4 - 2\sqrt{2}$$ $$	ext{LHS} = \frac{\sqrt{2} - 1}{4 - 2\sqrt{2}} = \frac{\sqrt{2} - 1}{2\sqrt{2}(\sqrt{2} - 1)} = \frac{1}{2\sqrt{2}} = \frac{\sqrt{2}}{4} = 	ext{RHS}$$ Hence proved.

Evaluate trigonometric expression given tangent value

Explanation

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