Prove each of the following identities (a) sec x + tan x =cos x/1 − sin x (b) tan^2 x/tan^2 x + 1= sin^2 x?

a)

(a) To prove $\sec(x) + \tan(x) = \frac{\cos(x)}{1 - \sin(x)}$:

Start with the left-hand side $\sec(x) + \tan(x)$: $\sec(x) + \tan(x) = \frac{1}{\cos(x)} + \frac{\sin(x)}{\cos(x)} = \frac{1 + \sin(x)}{\cos(x)}$

Now, let's simplify the right-hand side: $\frac{\cos(x)}{1 - \sin(x)} = \frac{\cos(x)}{1 - \sin(x)} \times \frac{1 + \sin(x)}{1 + \sin(x)} = \frac{\cos(x)(1 + \sin(x))}{1 - \sin^2(x)} = \frac{\cos(x) + \cos(x)\sin(x)}{\cos^2(x)} = \frac{1 + \sin(x)}{\cos(x)}$

Since both sides simplify to $\frac{1 + \sin(x)}{\cos(x)}$, the identity is proved.

(b) To prove $\frac{\tan^2(x)}{\tan^2(x) + 1} = \sin^2(x)$:

Start with the left-hand side $\frac{\tan^2(x)}{\tan^2(x) + 1}$: $\frac{\tan^2(x)}{\tan^2(x) + 1} = \frac{\sin^2(x)}{\sin^2(x) + \cos^2(x)}$

Now, use the trigonometric identity $\sin^2(x) + \cos^2(x) = 1$: $= \frac{\sin^2(x)}{1} = \sin^2(x)$

Hence, the identity is proved.