Question

Prove that. tan/1-cot+cot/1-tan=(1+sec cosec)

Answer 1

$\frac{tanx}{1-cotx}+\frac{cotx}{1-tanx}\\\\=\frac{tanx}{\frac{tanx-1}{tanx}}+\frac{1}{tanx}\frac{1}{1-tanx}\\\\=\frac{tan^2x}{tanx-1}-\frac{1}{tanx\ (1- tanx)}\\\\=\frac{tan^3x-1}{tanx(tanx-1)}\\\\=\frac{(tanx-1)(tan^2x+tanx+1)}{tanx(tanx-1)}\\\\if\ tanx\ \neq\ 1,\ then:\\\\=tanx+1+Cotx\\\\=1+\frac{sinx}{cosx}+\frac{cosx}{sinx}\\\\=1+\frac{sin^2+cos^2x}{sinx\ cosx}\\\\=1+secx\ cosecx$

Answer 2

Answer:

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Step-by-step explanation: