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serg [7]
3 years ago
9

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

Mathematics
1 answer:
lilavasa [31]3 years ago
6 0

Answer:

\frac{\sin^{6} x-\cos^{6}x }{1-\sin^{2}x \cos^{2}x  } =1-2\cos^{2} x proved.

Step-by-step explanation:

We have to prove that \frac{\sin^{6} x-\cos^{6}x }{1-\sin^{2}x \cos^{2}x  } =1-2\cos^{2} x

So, the left hand side = \frac{\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}

= \frac{(\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.

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