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Trigonometric Ratios Of Multiple And Sub Multiple Angles — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Ratios Of Multiple And Sub Multiple Angles3 Marks
Question
$\text{If} \tan\text{(A}+\text{B)}=\text{x}$ and $\tan\text{(A}-\text{B)}=\text{y},$find the values of $\tan2\text{A}$ and $\tan2\text{B}.$
✓

Answer

We have,
$\tan\text{(A}+\text{B)}=\tan\text{(A}-\text{B)}=\text{y}$
Now, $\tan2\text{A}=\tan\Big[\text{(A}+\text{B})+\text{(A}-\text{B})\Big]$
$=\frac{\tan\text{(A}+\text{B)+}\tan\text{(A}-\text{B)}}{1-\tan\text{(A}+\text{B)}\text{x}\tan\text{(A}-\text{B})}$
$=\frac{\text{x}+\text{y}}{\text{1}-\text{xy}}$
$\tan2\text{A}=\frac{\text{x}+\text{y}}{1-\text{xy}}$
Now, $\tan2\text{A}=\tan\Big[\text{(A}-\text{B)}-\text{(A}-\text{B)}\Big]$
$=\frac{\tan\text{(A}+\text{B)-}\tan\text{(A}-\text{B)}}{1+\tan\text{(A}+\text{B)}\text{x}\tan\text{(A}-\text{B})}$
$=\frac{\text{x}-\text{y}}{\text{1}+\text{xy}}$
$\therefore \tan2\text{B}=\frac{\text{x}-\text{y}}{\text{1}+\text{xy}}$

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