How do you prove $\frac{cot θ}{csc θ - 1} = \frac{csc θ + 1}{cot θ}$ $cottheta/(csctheta-1)=(csctheta+1)/cottheta$ ?

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Answer 1 · 2016-09-18T05:31:35

Please see below.

$\frac{cot θ}{csc θ - 1}$ $cottheta/(csctheta-1)$

= $\frac{cot θ}{csc θ - 1} \times \frac{csc θ + 1}{csc θ + 1}$ $cottheta/(csctheta-1)xx(csctheta+1)/(csctheta+1)$

= $\frac{cot θ (csc θ + 1)}{({csc}^{2} θ - 1)}$ $(cottheta(csctheta+1))/((csc^2theta-1)$

= $\frac{cot θ (csc θ + 1)}{{cot}^{2} θ}$ $(cottheta(csctheta+1))/cot^2theta$

= $\frac{csc θ + 1}{cot θ}$ $(csctheta+1)/cottheta$

How do you prove cotθcscθ−1=cscθ+1cotθcottheta/(csctheta-1)=(csctheta+1)/cottheta?