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Trigonometric Ratios of Multiple and Submultiple Angles — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSTrigonometric Ratios of Multiple and Submultiple Angles3 Marks
Question
If $\tan\text{A}=\text{x}\tan\text{B},$ prove that $\frac{\sin\text{(A}-\text{B)}}{\sin\text{(A}+\text{B)}}=\frac{\text{x}-\text{1}}{\text{x}+\text{1}}$
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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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