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
#SciPiPupil
0 Comments
You can let us know your questions in the comments section as well.