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

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Answer 1 · 2018-03-04T07:12:22

Given,

$(\frac{sin x}{csc x - 1}) + (\frac{sin x}{csc x + 1})$ $((sin x)/(cscx-1)) +((sinx)/(csc x+1))$

$= \frac{sin x (csc x + 1) + sin x (csc x - 1)}{(csc x + 1) (csc x - 1)}$ $=(sinx(cscx+1) + sinx(cscx-1))/((cscx+1)(cscx-1))$

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

$= \frac{sin x 2 csc x}{{cot}^{2} x}$ $=(sinx 2 cscx)/(cot^2 x)$ (as, ${csc}^{2} x - {cot}^{2} x = 1$ $csc^2 x - cot^2 x=1$ )

$= \frac{\frac{2 sin x}{sin x}}{{cot}^{2} x}$ $=((2 sinx)/(sinx))/cot^2 x$

$= \frac{2}{{cot}^{2} x}$ $=2/cot^2 x$

$= 2 {tan}^{2} x$ $=2 tan^2 x$

How do you prove (sinx/(cscx-1))+(sinx/(csc+1))=2tan^2x(sinxcscx−1)+(sinxcsc+1)=2tan2x(sinx/(cscx-1))+(sinx/(csc+1))=2tan^2x?