Question

If $\tan A=\sqrt{2}-1$, then prove that $\sin A\cos A=\frac{1}{2\sqrt{2}}$.

Answer 1

$sin2A=\left(\frac{2tanA}{1+ta{n}^{2}A}\right)=\left(\frac{2(\sqrt{2}-1)}{1+(\sqrt{2}-1{)}^2}\right)=\left(\frac{2(\sqrt{2}-1)}{1+2+1-2\sqrt{2}}\right)=\left(\frac{2(\sqrt{2}-1)}{4-2\sqrt{2}}\right)=\left(\frac{2(\sqrt{2}-1)}{2\sqrt{2}(\sqrt{2}-1)}\right)\therefore sin2A=\left(\frac{1}{\sqrt{2}}\right)or2sinAcosA=\left(\frac{1}{\sqrt{2}}\right)\therefore sinAcosA=\left(\frac{1}{2\sqrt{2}}\right)$