Question
If $\tan\text{A}=\frac{1-\cos\text{B}}{\sin\text{B}},$ then $\tan2\text{A}=\tan\text{B}$
✓
Answer
True.
Solution:
Given that, $\tan\text{A}=\frac{1-\cos\text{B}}{\sin\text{B}}=\frac{2\sin^2\frac{\text{B}}{2}}{2\sin\frac{\text{B}}{2}2\cos\frac{\text{B}}{2}}=\tan\frac{\text{B}}{2}$
$\tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}}=\frac{2\tan\frac{\text{B}}{2}}{1-\tan^2\frac{\text{B}}{2}}$
$\therefore2\text{A}=\tan\text{B}$
Hence, the statement is 'True'.
Solution:
Given that, $\tan\text{A}=\frac{1-\cos\text{B}}{\sin\text{B}}=\frac{2\sin^2\frac{\text{B}}{2}}{2\sin\frac{\text{B}}{2}2\cos\frac{\text{B}}{2}}=\tan\frac{\text{B}}{2}$
$\tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}}=\frac{2\tan\frac{\text{B}}{2}}{1-\tan^2\frac{\text{B}}{2}}$
$\therefore2\text{A}=\tan\text{B}$
Hence, the statement is 'True'.
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