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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
Question
Fill in the blank.
If $\tan\text{A}=\frac{1-\cos\text{B}}{\sin\text{B}},$ then $\tan2\text{A}=$ _______.
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Answer

If $\tan\text{A}=\frac{1-\cos\text{B}}{\sin\text{B}},$ then $\tan2\text{A}=$ $\tan\text{B}.$
Solution:
Given that, $\tan\text{A}=\frac{1-\cos}{\sin\text{B}}$
$\tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}}=\frac{2\Big(\frac{1-\cos\text{B}}{\sin\text{B}}\Big)}{1-\Big(\frac{1-\cos\text{B}}{\sin\text{B}}\Big)^2}$
$=\frac{2\Bigg(\frac{2\sin^2\frac{\text{B}}{2}}{2\sin\frac{\text{B}}{2}\cos\frac{\text{B}}{2}}\Bigg)}{1-\Bigg(\frac{2\sin^2\frac{\text{B}}{2}}{2\sin\frac{\text{B}}{2}\cos\frac{\text{B}}{2}}\Bigg)^2}$$\begin{bmatrix}\because1-\cos\text{B}=2\sin^2\frac{\text{B}}{2}\\\sin\text{B}=2\sin\frac{\text{B}}{2}\cos\frac{\text{B}}{2} \end{bmatrix}$
$=\frac{2\Bigg(\frac{\sin\frac{\text{B}}{2}}{\cos\frac{\text{B}}{2}}\Bigg)}{1-\Bigg(\frac{\sin\frac{\text{B}}{2}}{\cos\frac{\text{B}}{2}}\Bigg)^2}=\frac{2\tan\frac{\text{B}}{2}}{1-\tan^2\frac{\text{B}}{2}}=\tan\text{B}$
So, $\tan2\text{A}=\tan\text{B}$
Hence ,the value of the filler is $\tan\text{B}.$

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