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Mathematics 12 Online
OpenStudy (anonymous):

(tan theta / 1-cot theta) + (cot theta / 1-tan theta) =1 + tan theta + cot theta prove

OpenStudy (anonymous):

help any one

OpenStudy (anonymous):

gimme a sec

OpenStudy (anonymous):

plzzz

OpenStudy (anonymous):

ok \[\frac{tanx}{1-cotx} + \frac{cotx}{1-tanx} = \frac{tanx(1-tanx) + cotx(1-cotx)}{(1-tanx)(1-cotx)}\] \[=\frac{tanx - \tan^2x + cotx-\cot^2x}{(1-tanx)(1-cotx)}\]\[=\frac{tanx - \tan^2x + cotx-\cot^2x}{2 - cotx - tanx}\]\[=\frac{(tanx - \tan^2x + cotx-\cot^2x)+(2 - cotx - tanx) -(2 - cotx - tanx)}{(2 - cotx - tanx)}\]\[=1+ \frac{(tanx - \tan^2x + cotx-\cot^2x)-(2 - cotx - tanx)}{(2 - cotx - tanx)}\]\[=1+ \frac{(tanx - \tan^2x + cotx-\cot^2x)-2 + cotx + tanx)}{(2 - cotx - tanx)}\]\[=1+ \frac{(2tanx - \tan^2x + 2cotx-\cot^2x-2 )}{(2 - cotx - tanx)}\]\[=1+ \frac{(tanx(2 - cotx - tanx) +cotx(2-cotx-tanx) )}{(2 - cotx - tanx)}\]\[=1+ \frac{tanx(2 - cotx - tanx) }{(2 - cotx - tanx)} + \frac{cotx(2-cotx-tanx)}{(2-cotx-tanx)}\]\[= 1 + tanx + cotx\] QED

OpenStudy (anonymous):

THANKU YAAR

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