If tan ⁡ A = 1 − cos ⁡ B sin ⁡ B , prove that tan ⁡ 2 A = tan ⁡ B

Question

If   tan  ⁡  A  =                    1        −        cos        ⁡        B                    sin        ⁡        B            , prove that   tan  ⁡  2  A  =  tan  ⁡  B

Marlee Norman · Accepted Answer

You can proceed in this way:  tan  ⁡  A  =            1      −      cos      ⁡      B              sin      ⁡      B        =            2              sin        2            ⁡              B        2                    2      sin      ⁡              B        2            cos      ⁡              B        2              =            sin      ⁡              B        2                    cos      ⁡              B        2              =  tan  ⁡      B    2  And hence comparing, we can write that   A  =  n  π  +      B    2   where n is any integer.So we can say that   2  A  =  2  n  π  +  B  ⇒  tan  ⁡  2  A  =  tan  ⁡  (  2  n  π  +  B  )  =  tan  ⁡  BHence proved.

Emanuel Keith · Accepted Answer

You are on the right road.... Continuing calculation,

\begin{aligned} \frac{\frac{2 - 2 \cos B}{\sin B}}{1 - \frac{(1 - \cos B)^{2}}{\sin^{2} B}} & = \frac{\frac{2 - 2 \cos B}{\sin B}}{\frac{\sin^{2} B - 1 + 2 \cos B - \cos^{2} B}{\sin^{2} B}} \\ = \frac{\frac{2 - 2 \cos B}{\sin B}}{\frac{\sin^{2} B - \sin^{2} B - \cos^{2} B + 2 \cos B - \cos^{2} B}{\sin^{2} B}} \\ = \frac{\frac{2 - 2 \cos B}{\sin B}}{\frac{2 \cos B (1 - \cos B)}{\sin^{2} B}} \\ = \frac{2 \sin^{2} B (1 - \cos B)}{2 \cos B \sin B (1 - \cos B)} \\ = \frac{\sin B}{\cos B} \\ = \tan B \end{aligned}

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