Answers: 3
![\bf \cfrac{1}{cos(x)+1}+\cfrac{1}{cos(x)-1}=-2csc(x)cot(x)\\\\ -----------------------------\\\\ \cfrac{1}{cos(x)+1}+\cfrac{1}{cos(x)-1}\implies \cfrac{cos(x)-1+cos(x)+1}{[cos(x)+1][cos(x)-1]}\\\\ -----------------------------\\\\](/tpl/images/0363/6469/7e324.png)
![\bf \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 sin^2(\theta)=1-cos^2(\theta)\\\\ -----------------------------\\\\ \cfrac{cos(x)-1+cos(x)+1}{cos^2(x)-1}\implies \cfrac{cos(x)+cos(x)}{-[1-cos^2(x)]} \\\\\\ \cfrac{2cos(x)}{-sin^2(x)}\implies -2\cdot \cfrac{1}{sin(x)}\cdot \cfrac{cos(x)}{sin(x)}](/tpl/images/0363/6469/82850.png)
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