# 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)

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(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.

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