Question

tan x/
1 - cotx
+
cot x/

1-tan x
= sec x. Cosec x + 1​

Answer 1

Answer:

Step-by-step explanation:

$\dfrac{tanx}{1-cotx} +\dfrac{cotx}{1-tanx}=secx.cosecx+1\\ \\L.H.S.\\\\\dfrac{tanx}{1-cotx} +\dfrac{cotx}{1-tanx}\\\\=\dfrac{\dfrac{sinx}{cosx} }{1-\dfrac{cosx}{sinx} } +\dfrac{\dfrac{cosx}{sinx} }{1-\dfrac{sinx}{cosx} }\\\\\\=\dfrac{\dfrac{sinx}{cosx} }{\dfrac{sinx-cosx}{sinx} } +\dfrac{\dfrac{cosx}{sinx} }{\dfrac{cosx-sinx}{cosx} }\\\\\\={\dfrac{sinx}{cosx} }*{\dfrac{sinx}{sinx-cosx} }+{\dfrac{cosx}{sinx} }*{\dfrac{cosx}{-(sinx-cosx} )}$

$={\dfrac{sin^2x}{cosx(sinx-cosx)} }+{\dfrac{cos^2x}{sinx-(sinx-cosx)} }\\\\={\dfrac{sin^2x+cos^2x}{-cosxsinx(sinx-cosx)} }\\\\={\dfrac{sin^2x+cos^2x-(sinx-cosx)}{-cosxsinx(sinx-cosx)} }\\\\={\dfrac{cosxsinx(sinx-cosx)+(sinx-cosx)}{cosxsinx(sinx-cosx)} }\\\\=1+\dfrac{1}{cosx} \dfrac{1}{sinx}\\ \\=1+secx.cosecx$