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Verify the identity. quantity one minus sine of x divided by cosine of x equals cosine of x divided by quantity one plus sine of x

User Sridvijay
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now, recall that the conjugate of a + b, is just a - b, therefore, the conjugate of 1 - sin(x) is just 1 + sin(x).

so hmmm let's multiply the left-side by the conjugate of the numerator then.


\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta) \\\\\\ \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ -------------------------------\\\\


\bf \cfrac{1-sin(x)}{cos(x)}=\cfrac{cos(x)}{1+sin(x)} \\\\\\ \cfrac{1-sin(x)}{cos(x)}\cdot \cfrac{1+sin(x)}{1+sin(x)}\implies \cfrac{[1-sin(x)][1+sin(x)]}{cos(x)[1+sin(x)]} \\\\\\ \cfrac{1^2-sin^2(x)}{cos(x)[1+sin(x)]}\implies \cfrac{\underline{cos^2(x)}}{\underline{cos(x)}[1+sin(x)]}\implies \cfrac{cos(x)}{1+sin(x)}
User Pablonete
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