## Calculus: Early Transcendentals 8th Edition

(a) $$\sec^2 x=\frac{1}{\cos^2 x}$$ (b) $$\sec x\tan x=\frac{\sin x}{\cos^2 x}$$ (c) $$\cos x-\sin x=\frac{\cot x-1}{\csc x}$$
What we must do here is to differentiate both sides of the equation. (a) $$\tan x=\frac{\sin x}{\cos x}$$ First consider the left side. Its derivative would be $$A=(\tan x)'=\sec^2 x$$ Then consider the right side, using the Quotient Rule to differentiate. $$B=\Bigg(\frac{\sin x}{\cos x}\Bigg)'=\frac{(\sin x)'\cos x-\sin x(\cos x)'}{\cos^2 x}$$ $$B=\frac{\cos x\cos x-\sin x(-\sin x)}{\cos^2 x}$$ $$B=\frac{\cos^2 x+\sin^2 x}{\cos^2 x}$$ $$B=\frac{1}{\cos^2 x}$$ Therefore, the result would be $$\sec^2 x=\frac{1}{\cos^2 x}$$ (b) $$\sec x=\frac{1}{\cos x}$$ The left side: $$A=(\sec x)'=\sec x\tan x$$ The right side: we would use Reciprocal Rule. In detail, $$B=\Bigg(\frac{1}{\cos x}\Bigg)'=-\frac{(\cos x)'}{\cos^2 x}$$ $$B=\frac{\sin x}{\cos^2 x}$$ So, the result would be $$\sec x\tan x=\frac{\sin x}{\cos^2 x}$$ (c) $$\sin x+\cos x=\frac{1+\cot x}{\csc x}$$ The left side: $$A=(\sin x+\cos x)'=\cos x-\sin x$$ The right side (use Quotient Rule): $$B=(\frac{1+\cot x}{\csc x})'=\frac{(1+\cot x)'\csc x-(1+\cot x)(\csc x)'}{\csc^2 x}$$ $$B=\frac{-\csc^2 x\csc x-(1+\cot x)(-\csc x\cot x)}{\csc^2 x}$$ $$B=\frac{-\csc^3 x+\csc x\cot x+\csc x\cot^2 x}{\csc^2 x}$$ $$B=\frac{-\csc^2 x+\cot x+\cot^2 x}{\csc x}$$ $$B=\frac{-\csc^2+\cot x+(\csc^2-1)}{\csc x}$$ (since $\cot^2 x=\csc^2-1$)$$B=\frac{\cot x-1}{\csc x}$$ Eventually, we have $$\cos x-\sin x=\frac{\cot x-1}{\csc x}$$