![\bf \cfrac{1}{1+sin(x)}\cdot \cfrac{1-sin(x)}{1-sin(x)}\implies \cfrac{1-sin(x)}{[1+sin(x)][1-sin(x)]}\\\\ -----------------------------\\\\ recall\qquad \textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ and\qquad sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\ -----------------------------\\\\ thus \\\\\\ \cfrac{1-sin(x)}{1^2-sin^2(x)}\implies \cfrac{1-sin(x)}{1-sin^2(x)}\implies \cfrac{1-sin(x)}{cos^2(x)}](/tpl/images/0495/3682/e9977.png)
:include step by step in how to do this using trig identities!
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Mathematics, 03.02.2020 05:48 kelsotay5480
:include step by step in how to do this using trig identities!
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