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Rajan@2021
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If sec A = 17/8, verify that (3-4 sin^2A)/(4cos^2 A-3) =(3- tan^2 A )/(1-3tan^2 A)

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One of the most important and exam oriented question from trigonometric ratios in which we have given sec A = 17/8, and we asked to show that (3-4 sin^2A)/(4cos^2 A-3) =(3- tan^2 A )/(1-3tan^2 A)

RS Aggarwal, Class 10, chapter 10, question no 23

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