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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 Angles2 Marks
Question
If $\frac{\sin(\text{x}+\text{y})}{\sin(\text{x}-\text{y)}}=\frac{\text{a}+\text{b}}{\text{a}-\text{b}}$ show that $\frac{\tan\text{x}}{\tan\text{y}}=\frac{\text{a}}{\text{b}}.$
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Answer

$\frac{\sin\text{x}.\cos\text{y}+\sin\text{y}.\cos\text{x}}{\sin\text{x}.\cos\text{y}-\sin\text{y}.\cos\text{x}}=\frac{\text{a}+\text{b}}{\text{a}-\text{b}}$ $\Rightarrow\frac{\sin\text{x}.\cos\text{y}+\sin\text{y}.\cos\text{x}+\sin\text{x}.\cos\text{y}-\sin\text{y}.\cos\text{x}}{\sin\text{x}.\cos\text{y}+\sin\text{y}.\cos\text{x} -\sin\text{x}.\cos\text{y}+\sin\text{y}.\cos\text{x}}$ $=\frac{\text{a}+\text{b}+\text{a}-\text{b}}{\text{a}+\text{b}-\text{a}+\text{b}}$ $\Rightarrow\frac{2\sin\text{x}.\cos\text{y}}{2\sin\text{y}.\cos\text{x}}=\frac{\text{2a}}{\text{2b}}$ $\Rightarrow\frac{\tan\text{x}}{\tan\text{y}}=\frac{\text{a}}{\text{b}}$ Hence proved.

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