1.

Prove that \(\frac{sinA-cosA+1}{sinA+cosA-1}=\frac{1}{secA-tanA}\)

Answer»

\(\frac{sinA-cosA+1}{sinA+cosA-1}=\frac{1}{secA-tanA}\) 

L.H.S. divide numerator and denominator by cos A 

\(\frac{tanA-1+secA}{tanA+1-secA}\)

\(\frac{tanA-1+secA}{1-secA+tanA}\)

We know that 1 + tan2 A = sec

Or 1 = sec2 A - tan2 A = (sec A + tan A)(sec A – tan A) 

= \(\frac{secA+tanA-1}{(secA+tanA)(secA-tanA)-(secA-tanA)}\)

\(\frac{secA+tanA-1}{(secA-tanA)(secA+tanA-1}\)

\(\frac{1}{secA-tanA}\) , proved.



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