Question

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

Answer 1

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.