How do you prove $\frac{{cot}^{2} x}{1 + csc x} = \frac{1 - sin x}{sin x}$ $cot^2 x/ (1+csc x) = (1-sin x)/ sin x$ ?

Question

22431 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-08T10:32:07

As proved below.

${cot}^{2} x = {csc}^{2} x - 1, csc x = \frac{1}{sin x}$ $cot^2 x = csc^2 x - 1, csc x = 1/ sin x$

$\frac{{cot}^{2} x}{1 + csc x} = \frac{{csc}^{2} x - 1}{csc x + 1}$ $cot^2 x / (1 + csc x) = (csc^2 x - 1) / (csc x + 1)$

$=> (cancel(csc x + 1) (csc x - 1)) / cancel(csc x + 1)$

$=> csc x - 1 = (1/sin x - 1) color(violet)(= (1-sin x) / sin x$

How do you prove cot^2 x/ (1+csc x) = (1-sin x)/ sin xcot2x1+cscx=1−sinxsinxcot^2 x/ (1+csc x) = (1-sin x)/ sin x?