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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
If $\tan\text{A}=\text{x}\tan\text{B},$ prove that $\frac{\sin\text{(A}-\text{B)}}{\sin\text{(A}+\text{B)}}=\frac{\text{x}-\text{1}}{\text{x}+\text{1}}$
✓

Answer

We have,
$\tan\text{A}=\text{x}\tan\text{B}$
$\frac{\sin\text{A}}{\cos\text{B}}=\text{x}\frac{\sin\text{B}}{\cos\text{B}}$ $\Big[\because\tan\theta=\frac{\sin\theta}{\cos\theta}\Big]$
$\Rightarrow\sin\text{A}\cos\text{B}=\text{x}\cos\text{A}\sin\text{B} ...(1)$
Now, $\frac{\sin\text{(A}-\text{B)}}{\sin\text{(A}+\text{B)}}=\frac{\sin\text{A}\cos\text{B}-\sin\text{B}\cos\text{A}}{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}} $
$=\frac{\text{x}\cos\text{A}\sin\text{B}-\cos\text{A}\sin\text{B}}{\text{x}\cos\text{A}\sin\text{B}+\cos\text{A}\sin\text{B}} $ $[$Using equation$]$
$=\frac{\cos\text{A}\sin\text{B}\text{(x}-{1)}}{\cos\text{A}\sin\text{B}\text{(x}+{1)}}$
$=\frac{\text{x}-1}{\text{x}+1}$
$\therefore\frac{\sin\text{(A}-\text{B)}}{\sin\text{(A}+\text{B)}}=\frac{\text{x}-1}{\text{x}+1}$
Hence proved.

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