Question

(Tanx/1-cotx)+(cotx/1-tanx)=1+secx*cscx

Answer 1

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

Answer 2

\[\frac{ 1 }{ 1-\cot x } (\frac{ 1 }{ \cot x } - \cot ^2x) =\\\frac{ 1 }{ 1-\cot x } (\frac{ 1 }{ \cot x } ) (1-\cot ^3x) =\\\frac{ 1 }{ 1-\cot x } (\frac{ 1 }{ \cot x } ) (1-\cot x)(1+\cot x +\cot ^2x) =\\\]

\[ \[(\frac{ 1 }{ \cot x } ) (1+\cot x +\cot ^2x) =\\ \frac{ 1 }{ \cot x }+\frac{ \cot x }{ \cot x } +\frac{ \cot ^2x }{ \cot x }=\\tanx + 1+ \cot x=\\1+(tanx +\cot x)=\\\]\\]

Answer 3

now it is sufficient to write tan x , cot x in sinx , cos x

Answer 4

\[1+\tan x + \cot x = 1 +\frac{ sinx }{ \cos x }+\frac{ \cos x }{\sin x } =\\1+ \frac{ \sin^2x +\cos ^2x }{ \sin x \cos x }=\\1 +\frac{ 1 }{ sinx \cos x }= 1+\frac{ 1 }{ sinx }\frac{ 1 }{ \cos x }=\\1+\sec x \times \csc x \]