Respuesta :
Answer:
It's an identity
Step-by-step explanation:
[tex]\sin^4x-\sin^2x=\cos^4x-\cos^2x\\\\L_s=\sin^4x-\sin^2x=\sin^2x(\sin^2x-1)=\sin^2x(-\cos^2x)=-\sin^2x\cos^2x\\\\R_s=\cos^4x-\cos^2x=\cos^2x(\cos^2x-1)=\cos^2x(-\sin^2x)=-\sin^2x\cos^2x\\\\L_s=R_s\qquad\bold{CORRECT}[/tex]
[tex]\text{Used:}\\\\\sin^2x+\cos^2x=1\Rightarrow\left\{\begin{array}{ccc}\cos^2x=1-\sin^2x\\\sin^2x=1-\cos^2x\end{array}\right\Rightarrow\left\{\begin{array}{ccc}-\cos^2x=\sin^2x-1\\-\sin^2x=\cos^2x-1\end{array}\right[/tex]
Step-by-step explanation:
L.H.S
= sin⁴x − sin²x
= sin²x ( sin²x - 1 )
=(1-cos²x) (1-cos²x-1) [ Trigonometric formula " sin²x=1-cos²x " ]
= (1-cos²x) (-cos²x)
=-cos²x+ cos⁴x
= cos⁴x - cos²x
=R.H.S