Question

# $$\frac{tanA+secA-1}{tanA-secA+1}=\frac{1+sinA}{cosA}$$

Solution

## $$\cfrac{\tan{A} + \sec{A} - 1}{\tan{A} - \sec{A} + 1} = \cfrac{1 + \sin{A}}{\cos{A}}$$Taking L.H.S.-$$\cfrac{\tan{A} + \sec{A} - 1}{\tan{A} - \sec{A} + 1}$$$$= \cfrac{\left( \tan{A} + \sec{A} \right) - \left( \sec^{2}{A} - \tan^{2}{A} \right)}{\tan{A} - \sec{A} + 1} \; \left[ \because 1 + \tan^{2}{A} = \sec^{2}{A} \right]$$$$= \cfrac{\left( \tan{A} + \sec{A} \right) - \left( \sec{A} + \tan{A} \right) \left( \sec{A} - \tan{A} \right)}{\tan{A} - \sec{A} + 1}$$$$= \cfrac{\left( \tan{A} + \sec{A} \right) \left( 1 - \left( \sec{A} - \tan{A} \right) \right)}{\tan{A} - \sec{A} + 1}$$$$= \cfrac{\left( \tan{A} + \sec{A} \right) \left( 1 - \sec{A} + \tan{A} \right)}{\tan{A} - \sec{A} + 1}$$$$= \tan{A} + \sec{A}$$$$= \cfrac{\sin{A}}{\cos{A}} + \cfrac{1}{\cos{A}}$$$$= \cfrac{1 + \sin{A}}{\cos{A}}$$$$=$$ R.H.S.Hence proved.Mathematics

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