Answer:
[tex]\frac{\sin^{6} x-\cos^{6}x }{1-\sin^{2}x \cos^{2}x } =1-2\cos^{2} x[/tex] proved.
Step-by-step explanation:
We have to prove that [tex]\frac{\sin^{6} x-\cos^{6}x }{1-\sin^{2}x \cos^{2}x } =1-2\cos^{2} x[/tex]
So, the left hand side = [tex]\frac{\sin^{6} x-\cos^{6}x }{1-\sin^{2}x \cos^{2}x }[/tex]
= [tex]\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 }}[/tex] {Since we have the formula [tex]a^{3} -b^{3}= (a-b) (a^{2} +ab+b^{2} )[/tex]}
= [tex]\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 }}[/tex] {Since we have the formula [tex]a^{2}+b^{2} = (a+b)^{2} -2ab[/tex]}
= [tex]\frac{(\sin^{2}x-\cos^{2} x )(1-\sin^{2}x \cos^{2}x)}{(1-\sin^{2}x \cos^{2}x)}[/tex]
= [tex](\sin^{2}x-\cos^{2} x )[/tex]
= [tex]1-2\cos^{2} x[/tex] {Since [tex]\sin^{2}x =1-\cos^{2} x[/tex]}
= Right hand side
Hence, proved.
