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Tan∅/1-cot∅ + cot∅/1-tan∅ = 1+ sec∅ cosec∅ |
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Answer» Step-by-step explanation: \begin{gathered}\frac{tanx}{1-cotx}+\frac{cotx}{1-tanx}\\\\=\frac{tanx}{\frac{tanx-1}{tanx}}+\frac{1}{tanx}\frac{1}{1-tanx}\\\\=\frac{TAN^2x}{tanx-1}-\frac{1}{tanx\ (1- tanx)}\\\\=\frac{tan^3x-1}{tanx(tanx-1)}\\\\=\frac{(tanx-1)(tan^2x+tanx+1)}{tanx(tanx-1)}\\\\if\ tanx\ \neq\ 1,\ then:\\\\=tanx+1+Cotx\\\\=1+\frac{sinx}{COSX}+\frac{cosx}{sinx}\\\\=1+\frac{sin^2+COS^2x}{sinx\ cosx}\\\\=1+secx\ COSECX\end{gathered} 1−cotx tanx
+ 1−tanx cotx
= tanx tanx−1
tanx
+ tanx 1
1−tanx 1
= tanx−1 tan 2 x
− tanx (1−tanx) 1
= tanx(tanx−1) tan 3 x−1
= tanx(tanx−1) (tanx−1)(tan 2 x+tanx+1)
if tanx
= 1, then: =tanx+1+Cotx =1+ cosx sinx
+ sinx cosx
=1+ sinx cosx sin 2 +cos 2 x
=1+secx cosecx
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