Verify the following identity and indicate where the equality is valid:
$$\dfrac{\cos^2 x}{1-\sin x}=1+\sin x $$

Question

Verify the following identity and indicate where the equality is valid:
$$\dfrac{\cos^2 x}{1-\sin x}=1+\sin x  $$

Uniteasy Admin · Accepted Answer

By using the Pythagorean identity, we can get
$$\cos^2 x = 1- \sin^2 x$$
and then the factorization
$$ 1- \sin^2 x = (1-\sin x)(1+\sin x)$$
the following sequence of equalities can
be established:
$$\dfrac{\cos^2 x}{1-\sin x} $$
$$=\dfrac{1- \sin^2 x}{1- \sin x} $$
$$=\dfrac{ (1-\sin x)(1+\sin x)}{1- \sin x} $$
$$=1+\sin x $$
The identity is valid as long as$$ 1− \sin x {=}\mathllap{/\,} 0 $$ or $$ \sin x {=}\mathllap{/\,} 1$$
Based on the properties of sine function, 
when $$ x {=}\mathllap{/\,} \dfrac{\pi}{2}+2k\pi $$ where k denotes any integer, the identity is valid

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