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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