Question: Prove that: $\dfrac{tan x}{secx -1} - \dfrac{sinx}{1 + cosx} = 2cotx$.


Solution:

To prove: $\dfrac{tan x}{secx -1} - \dfrac{sinx}{1 + cosx} = 2cotx$

Taking LHS

$= \dfrac{tan x}{secx -1} - \dfrac{sinx}{1 + cosx}$

[$tanx = \frac{sinx}{cosx}$ and $secx= \frac{1}{cosx}$]

$= \dfrac{ \frac{sinx}{cosx} }{ \frac{1}{cosx} -1} - \dfrac{sinx }{1 + cosx}$

$= \dfrac{\frac{sinx}{cosx}}{\frac{1 -cosx}{cosx}} - \dfrac{sinx}{1 + cosx}$

$= \dfrac{sinx}{cosx} × \dfrac{cosx}{1 -cosx} - \dfrac{sinx}{1 + cosx}$

$= \dfrac{sinx}{1-cosx} - \dfrac{sinx}{1+cosx}$

[Taking sinx common from both terms]

$= sinx \left ( \dfrac{1}{1 - cosx} - \dfrac{1}{1 + cosx} \right )$

$= sinx \left ( \dfrac{(1 +cosx )- (1- cosx)}{(1+cosx)(1-cosx)}  \right )$

$= sinx \left ( \dfrac{1+ cosx -1 +cosx}{1 - cos^2x} \right)$

[Using, 1 - cos²x = sin²x]

$= sinx \left ( \dfrac{2cosx}{sin^2x} \right)$

$= \dfrac{2cosx}{sinx}$

$= 2cotx$

= RHS
#proved


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