Question

Sin ^6x-cos^6÷1-sin^2x*cos^2x=1-2cos^2x

​

Answer 1

`(\sin^(6) x-\cos^(6)x )/(1-\sin^(2)x \cos^(2)x  ) =1-2\cos^(2) x` proved.

Explanation:

We have to prove that
`(\sin^(6) x-\cos^(6)x )/(1-\sin^(2)x \cos^(2)x  ) =1-2\cos^(2) x`

So, the left hand side =
`(\sin^(6) x-\cos^(6)x )/(1-\sin^(2)x \cos^(2)x  )`

=
`\frac{(\sin^(2)x-\cos^(2) x )(\sin^(4) x+\cos^(4)x+\sin^(2)x \cos^(2)x)   }{{1-\sin^(2)x \cos^(2)x  }}` {Since we have the formula
`a^(3) -b^(3)= (a-b) (a^(2)  +ab+b^(2) )`}

=
`\frac{(\sin^(2)x-\cos^(2) x )[(\sin^(2)x+\cos^(2)x  )^(2)-2\sin^(2)x\cos^(2) x+ \sin^(2)x\cos^(2) x ]}{{1-\sin^(2)x \cos^(2)x  }}` {Since we have the formula
`a^(2)+b^(2) = (a+b)^(2) -2ab`}

=
`((\sin^(2)x-\cos^(2) x )(1-\sin^(2)x \cos^(2)x))/((1-\sin^(2)x \cos^(2)x))`

=
`(\sin^(2)x-\cos^(2) x )`

=
`1-2\cos^(2) x` {Since
`\sin^(2)x =1-\cos^(2) x`}

= Right hand side

Hence, proved.