Question

Verify each identity:
(1+cost)/(1-cost)=(csct+cot)^2

Answer 1

multiply top and bottom of the left hand side by the conjugate of 1-cost

Answer 2

fyi, the "conjugate" just means you change the - to a +, so multiply top and bottom by
1+cost

Answer 3

\(\bf \cfrac{1+cos(t)}{1-cos(t)}=[csc(t)+cot(t)]^2\\ \quad \\ \quad \\
\cfrac{1+cos(t)}{1-cos(t)}\cdot \cfrac{1+cos(t)}{1+cos(t)}\implies
\cfrac{[1+cos(t)][1+cos(t)]}{[1-cos(t)][1+cos(t)]}\\ \quad \\
\textit{recall that }{\color{blue}{ (a-b)(a+b) = a^2-b^2}}\qquad thus\\ \quad \\
\cfrac{[1+cos(t)][1+cos(t)]}{[1-cos(t)][1+cos(t)]}\implies \cfrac{[1+cos(t)]^2}{{\color{blue}{ 1^2-cos^2(t)}}}\\ \quad \\

\textit{recall that }sin^2(\theta)+cos^2(\theta)=1\implies {\color{blue}{ sin^2(\theta)=1-cos^2(\theta)}}\\ \quad \\
\cfrac{[1+cos(t)]^2}{1^2-cos^2(t)}\implies \cfrac{[1+cos(t)]^2}{1-cos^2(t)}\implies \cfrac{[1+cos(t)]^2}{{\color{blue}{ sin^2(t)}}}\)

expand the numerator, and distribute the denominator, see what you get

Answer 4

Thanks! :)

Answer 5

yw