Answer:
TRUE
Step-by-step explanation:
[tex]\dfrac{\sin^2(x)+\cos(x)+\cos^2(x)-1}{\sec(x)}=\dfrac{(\sin^2(x)+\cos^2(x))-1+\cos(x)}{\sec(x)}\\\\\text{use}\ \sin^2x+\cos^2x=1\ \text{and}\ \sec x=\dfrac{1}{\cos x}\\\\=\dfrac{1-1+\cos(x)}{\frac{1}{\cos (x)}}=\dfrac{\cos(x)}{\frac{1}{\cos(x)}}=\cos(x)\cdot\dfrac{\cos(x)}{1}=\cos^2(x)[/tex]
