Answer:
Trigonometric Identities used:
[tex]\sin^2 \theta + \cos^2 \theta \equiv 1[/tex]
[tex]\sec \theta \equiv \dfrac{1}{\cos \theta}[/tex]
[tex]\cot \theta \equiv \dfrac{1}{\tan \theta}[/tex]
[tex]\sec^2 \theta \equiv 1 + \tan^2 \theta[/tex]
[tex]\sin(\theta)=\cos \left(\dfrac{\pi}{2}-\theta\right)[/tex]
Part (a)
[tex](1+ \cos(x))(1-\cos(x))[/tex]
[tex]=1-\cos(x)+\cos(x)-\cos^2(x)[/tex]
[tex]=1-\cos^2(x)[/tex]
[tex]= \sin^2(x)[/tex]
Part (b)
[tex]\dfrac{1}{\cot^2(x)}-\dfrac{1}{\cos^2(x)}[/tex]
[tex]=\tan^2(x)-\sec^2(x)[/tex]
[tex]=\tan^2(x)-(1 + \tan^2(x))[/tex]
[tex]=\tan^2(x)-1 - \tan^2(x)[/tex]
[tex]=-1[/tex]
Part (c)
[tex]\sec^2\left(\dfrac{ \pi }{2}-x\right) \left[\sin^2(x)-\sin^4(x) \right][/tex]
[tex]=\dfrac{1}{\cos^2\left(\dfrac{ \pi }{2}-x\right)} \left[\sin^2(x)-\sin^4(x) \right][/tex]
[tex]=\dfrac{1}{\sin^2(x)} \left[\sin^2(x)-\sin^4(x) \right][/tex]
[tex]= \dfrac{\sin^2(x)-\sin^4(x)}{\sin^2(x)}[/tex]
[tex]= \dfrac{\sin^2(x)}{\sin^2(x)} - \dfrac{\sin^4(x)}{\sin^2(x)}[/tex]
[tex]= \dfrac{\sin^2(x)}{\sin^2(x)} - \dfrac{(\sin^2(x))(\sin^2(x))}{\sin^2(x)}[/tex]
[tex]=1- \sin^2(x)[/tex]
[tex]= \cos^2(x)[/tex]
