Prove that 2sec^2x-2sec^2xsinx-sin^2x-cos^2x=1.

2sec² (x) - 2sec² (x) sin (x) - sin² (x) - cos² (x) = 1

2sec² (x) - 2sec² (x) sin (x) - sin² (x) - cos² (x) = sin² (x) + cos² (x)

+ sin² (x) + cos² (x) + sin² (x) + cos² (x)

2sec² (x) - 2sec² (x) sin (x) = 2sin² (x) + 2cos² (x)

2[sec² (x) ] - 2[sec² (x) sin (x) ] = 2[sin² (x) + cos² (x) ]

2[sec² (x) - sec² (x) sin (x) ] = 2 (1)

2 2

sec² (x) - sec² (x) sin (x) = 1

sec² (x) sec² (x)

1 - sin (x) = cos² (x)

sin² (x) + cos² (x) - sin (x) = cos² (x)

- cos² (x) - cos² (x)

sin² (x) - sin (x) = 0

sin (x) [sin (x) ] - sin (x) [1] = 0

sin (x) [sin (x) - 1] = 0

sin (x) = 0 or sin (x) - 1 = 0

sin⁻¹[sin (x) ] = sin⁻¹ (0) + 1 + 1

x = 0 sin (x) = 1

sin⁻¹[sin (x) ] = sin⁻¹ (1)

x ≈ 1.5707

The solution is actually equal to zero.