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If cotA = \(\frac{4}5\) , prove that \(\frac{(sin\,A+cos\,A)}{(sin\,A-cos\,A)}\) = 9 |
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Answer» Given that cot A = \(\frac{4}5\), Now,\(\frac{sin\,A+cos\,A}{sin\,A-cos\,A}\) = \(\frac{1+\frac{cosA}{sinA}}{1-\frac{cosA}{sinA}}\) (Dividing both numerator and denominator by sin A) = \(\frac{1+cotA}{1-cotA}\) = \(\frac{1+\frac{4}{5}}{1-\frac{4}{5}}\) = \(\frac{\frac{9}{5}}{\frac{1}{5}}\) = 9 (∵ cot A = 4/5) Hence Proved |
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