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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 Angles4 Marks
Question
If $ \tan\text{A}+\tan\text{B}=\text{a}$ and $\cot\text{A}+\cot\text{B}=\text{b},$ prove that $\cot\text{(A}+\text{B)}=\frac{1}{\text{a}}-\frac{1}{\text{b}}.$
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Answer

We have, $\tan\text{A}+\tan\text{B}=\text{a}$ $\text{and}\ \cot\text{A}+\cot\text{B}=\text{b}$ $\cot\text{A}+\cot\text{B}=\text{b}$ $\Rightarrow\frac{1}{\tan\text{A}}+\frac{1}{\tan\text{B}}=\text{b}$ $\Big[\because\cot\theta=\frac{1}{\tan\theta}\Big]$ $\Rightarrow\frac{\tan\text{B}+\tan\text{A}}{\tan\text{A}\tan\text{B}}=\text{b}$ $\big[\because\tan\text{A}+\tan\text{B}=\text{a}\big]$ $\Rightarrow\frac{\text{a}}{\tan\text{A}\tan\text{B}}=\text{b}$ $\Rightarrow\frac{\text{a}}{\text{b}}=\tan\text{A}\tan\text{B}$ $\because\cot\text{(A}+\text{B)}=\frac{1}{\tan\text{(A}+\text{B)}}$ $=\frac{\frac{1}{\tan\text{A}+\tan\text{B}}}{1-\tan\text{A}\tan\text{B}}$ $=\frac{1-\tan\text{A}\tan\text{B}}{\tan\text{A} +\tan\text{B}}$ $=\frac{1-\frac{\text{a}}{\text{b}}}{\text{a}}$ $=\frac{\text{b}-\text{a}}{\text{ab}}$ $\big[\because\tan\text{A}+\tan\text{B}=\frac{\text{a}}{\text{b}}\big]$ $=\frac{\text{b}}{\text{ab}}-\frac{\text{a}}{\text{ab}}$ $=\frac{\text{1}}{\text{a}}-\frac{\text{1}}{\text{b}}$ $\therefore\cot\text{(A}+\text{B)}=\frac{1}{\text{a}}-\frac{1}{\text{b}}$ Hence proved.

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