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Prove that \(\frac{sinA-cosA+1}{sinA+cosA-1}=\frac{1}{secA-tanA}\) |
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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 = sec2 A 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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