Question

Verify the following identity and indicate where the equality is valid:

\dfrac{\cos^2 x}{1-\sin x}=1+\sin x

Collected in the board: Trigonometry

Steven Zheng posted 2 months ago


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

Steven Zheng posted 2 months ago

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