Here, we can prove the Right hand side (RHS) by solving the Left hand side (LHS),

We have,

LHS :

$\sf\dfrac{sinA - cosA + 1}{sinA + cosA - 1} = \dfrac{1}{secA - tanA} $

Now, solving LHS (Left Hand Side),

• Dividing both Numerator and Denominator on LHS by cosA,

$ \sf \dfrac{ \dfrac{sinA}{cosA} - \dfrac{cosA}{cosA} + \dfrac{1}{cosA}}{ \dfrac{sinA}{cosA} + \dfrac{cosA}{cosA} - \dfrac{1}{cosA}} $

$ \to \sf \dfrac{tanA - 1 + secA}{tanA + 1 - secA} $

We know that, we can also write 1 as sec²A - tan²A, so, replacing 1 by sec²A - tan²A on the denominator in LHS,

$ \to \sf \dfrac{secA + tanA - 1}{(secA + tanA)(secA - tanA) - (secA - tanA)}$

$ \to \sf \dfrac{secA + tanA - 1}{(secA - tanA)(secA + tanA - 1)} $

Cancelling, secA + tanA - 1 from the numerator and the denominator as they are same units,

We get,

$ \to \sf \dfrac{\cancel{secA + tanA - 1}}{(secA - tanA)\cancel{(secA + tanA - 1)}} $

$ \to \sf \dfrac{1}{secA - tanA}$

: RHS

• Hence Proved!!

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