secA+tanA-1/tanA-secA+1= cosA/1-sinA - Maths - Introduction to Trigonometry

Question

secA+tanA-1/tanA-secA+1= cosA/1-sinA

Ankush Jain · Accepted Answer

To prove: $\frac{\sec A + \tan A - 1}{\tan A - \sec A + 1} = \frac{\cos A}{1 - \sin A}$ Proof: LHS = $\frac{\sec A + \tan A - 1}{\tan A - \sec A + 1} = \frac{\sec A + \tan A - (\sec^{2} A - \tan^{2} A)}{\tan A - \sec A + 1} = \frac{(\sec A + \tan A) - (\sec A + \tan A) (\sec A - \tan A)}{\tan A - \sec A + 1} = \frac{(\sec A + \tan A) [1 - (\sec A - \tan A)]}{\tan A - \sec A + 1} = \frac{(\sec A + \tan A) [1 - \sec A + \tan A]}{\tan A - \sec A + 1} = \sec A + \tan A = \frac{1}{\cos A} + \frac{\sin A}{\cos A} = \frac{1 + \sin A}{\cos A} = \frac{1 + \sin A}{\cos A} \times \frac{1 - \sin A}{1 - \sin A} = \frac{1 - \sin^{2} A}{\cos A (1 - \sin A)} = \frac{\cos^{2} A}{\cos A (1 - \sin A)} = \frac{\cos A}{1 - \sin A} = R H S [H e n c e p r o v e d]$