CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Compound Angles1 Mark
Question
If $\tan\text{A}=\frac{1-\cos\text{B}}{\sin\text{B}},$ then find value of $\tan2\text{A}.$
✓
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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