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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
Prove that: $\frac{\tan\text{A}+\tan\text{B}}{\tan\text{A}-\tan\text{B}}=\frac {\sin\text{A+B}}{\sin\text{A-B}}$
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Answer

$\text{L.H.S}=$ $\frac{\tan\text{A}+\tan\text{B}}{\tan\text{A}-\tan\text{B} }$ $=\frac{\frac{\sin\text{A}}{\cos\text{A}}+\frac{\sin\text{B}}{\cos\text{B}}}{\frac{\sin\text{A}}{\cos\text{A}}-\frac{\sin\text{B}}{\cos\text{B}}}$ $\Big[\because\tan\theta=\frac{\sin\theta}{\cos\theta}\Big]$ $=\frac{\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}}{{\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}}{\cos\text{A}\cos\text{B}}}{}}$ $=\frac{\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}}{\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}}$ $\begin{bmatrix}\because\sin\text{(A+B)}=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B}\\\text{and,}\sin\text{(A-B)}=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B} \end{bmatrix}$ $=\frac{\sin\text{(A+B)}}{\sin\text{(A-B)}}$ $\therefore\frac{\tan\text{A}+\tan\text{B}}{\tan\text{A}-\tan\text{B}}=\frac{\sin\text{(A+B)}}{\sin\text{(A-B)}} $ $=\text{R.H.S}$ $\text{L.H.S}=\text{R.H.S}$ Hence proved.

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