If = 1- , then find value of 2A.

Question

If $\tan\text{A}=\frac{1-\cos\text{B}}{\sin\text{B}},$ then find value of $\tan2\text{A}.$

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Answer

given $\tan\text{A}=\frac{1\cos\text{B}}{\sin\text{B}}=\frac{2\sin^2\frac{\text{B}}{2}}{2\sin\frac{\text{B}}{2}\cos\frac{\text{B}}{2}}=\tan\frac{\text{B}}{2}$ $\tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}}$ Subtitute the value of tan A, we get $\tan2\text{A}=\frac{2\tan\frac{\text{B}}{2}}{1-\tan^2\frac{\text{B}}{2}}=\tan\text{B}$

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