If `tanA=sqrt(2)-1` show that `sinAcosA=(sqrt(2))/4`

Explanation

`tan A=sqrt{2}-1`

we know `tan A=frac{P}{B}=frac{sqrt{2}-1}{1}`


`therefore H=sqrt{(sqrt{2}-1)^{2}+1^{2}}=sqrt{4-2 sqrt{2}} `

`therefore sin A=frac{sqrt{2}-1}{sqrt{4-2 sqrt{2}}} quad cos A=frac{1}{sqrt{4-2 sqrt{2}}} `

`therefore sin A cos A=frac{sqrt{2}-1}{sqrt{4}-2 sqrt{2}} times frac{1}{sqrt{4-2 sqrt{2}}}=frac{sqrt{2}-1}{4-2 sqrt{2}} `

`=frac{sqrt{2}-1}{4-2 sqrt{2}} times frac{4+2 sqrt{2}}{4+2 sqrt{2}} `

`=frac{4 sqrt{2}+4-4-2 sqrt{2}}{16-8}=frac{2 sqrt{2}}{8}=frac{sqrt{2}}{4}`