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Prove: $$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}=\frac{1+\sin A}{\cos A}$$

My attempt:

LHS= $$\frac{\tan A+\sec A-1}{\tan A-\sec A+1}$$ $$=\frac{\frac{\sin A}{\cos A}+\frac{1}{\cos A}-1}{\frac{\sin A}{\cos A}-\frac{1}{\cos A}+1}$$ $$=\frac{\sin A+1-\cos A}{\sin A-1+\cos A}$$ $$\text{Using componendo and dividendo}$$ $$\frac{\sin A+1-\cos A+\sin A-1+\cos A}{\sin A+1-\cos A-\sin A+1-\cos A}$$ $$\frac{2\sin A}{2-2\cos A}=\frac{\sin A}{1-\cos A}=\frac{\sin A(1+\cos A)}{(1-\cos A)(1+\cos A)}$$ $$\frac{\sin A(1+\cos A)}{\sin^2 A}=\frac{1+\cos A}{\sin A}$$ $$\text{Which is not equal to right hand side!}$$

Have I done something wrong in the componendo dividendo step? I dont know how to use componendo dividendo rule . I saw it being used like this in some question and hence applied it here the same way. Maybe I am wrong in application of that rule. Please tell me the right way to use it. Thank you.

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marked as duplicate by lab bhattacharjee trigonometry Jun 13 '17 at 16:29

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Note \begin{eqnarray} LHS&=&\frac{\tan A+\sec A-1}{\tan A-\sec A+1}\\ &=&\frac{\frac{\sin A}{\cos A}+\frac{1}{\cos A}-1}{\frac{\sin A}{\cos A}-\frac{1}{\cos A}+1}\\ &=&\frac{\sin A+1-\cos A}{\sin A-1+\cos A}\\ &=&\frac{\sin A+1-\cos A}{\sin A-1+\cos A}\cdot\frac{\sin A+1+\cos A}{\sin A+1+\cos A}\\ &=&\frac{(\sin A+1)^2-\cos^2 A}{(\sin A+\cos A)^2-1}\\ &=&\frac{\sin^2 A-\cos^2 A+2\sin A+1}{2\sin A\cos A}\\ &=&\frac{2\sin^2 A+2\sin A}{2\sin A\cos A}\\ &=&\frac{1+\sin A}{\cos A}\\ &=&RHS \end{eqnarray}

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