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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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