Question

Verify sin^4x − sin^2x = cos^4x − cos^2x is an identity

Answer 1

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          

Answer 2

Answer:

It's an identity

Step-by-step explanation:

$\sin^4x-\sin^2x=\cos^4x-\cos^2x\\\\L_s=\sin^4x-\sin^2x=\sin^2x(\sin^2x-1)=\sin^2x(-\cos^2x)=-\sin^2x\cos^2x\\\\R_s=\cos^4x-\cos^2x=\cos^2x(\cos^2x-1)=\cos^2x(-\sin^2x)=-\sin^2x\cos^2x\\\\L_s=R_s\qquad\bold{CORRECT}$

$\text{Used:}\\\\\sin^2x+\cos^2x=1\Rightarrow\left\{\begin{array}{ccc}\cos^2x=1-\sin^2x\\\sin^2x=1-\cos^2x\end{array}\right\Rightarrow\left\{\begin{array}{ccc}-\cos^2x=\sin^2x-1\\-\sin^2x=\cos^2x-1\end{array}\right$