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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 $\tan\text{A}=\frac{\text{m}}{\text{m-1}}$ and $\tan\text{B}=\frac{1}{\text{2m-1}},$ then prove that $\text{A-B}=\frac{\pi}{4}$
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Answer

We have, $\tan\text{A}=\frac{\text{m}}{\text{m-1}}$ and $\tan\text{B}=\frac{\text{1}}{2\text{m}-1}$ Now, $\tan\text{(A-B)}=\frac{\tan\text{A}-\tan\text{B}}{1+\tan\text{A}{\tan\text{B}}}$ $=\frac{\frac{\text{m}}{\text{m-1}}-\frac{1}{{\text{2m}-1}}}{1+\frac{\text{m}}{\text{m-1}}\times\frac{1}{\text{2m-1}}}$ $=\frac{\frac{\text{m(2m-1)}-(\text{m-1})}{(\text{m-1})(\text{2m-1})}}{1+\frac{\text{m}}{\text{(m-1)}\text{(2m-1)}}}$ $=\frac{\frac{\text{m(2m-1)}-(\text{m-1})}{(\text{m-1})(\text{2m-1})}}{\frac{\text{(m-1)}\text{(2m-1)}+\text{(m)}}{\text{(m-1)}\text{(2m-1)}}}$ $=\frac{\text{m}(\text{2m-1)}-\text{(m-1)}}{\text{(m-1)}\text{(2m-1)}+\text{(m)}}$ $=\frac{\text{2m}^2-\text{m}-\text{m}+1}{\text{2m}^2-\text{m}-\text{2m}+1+\text{m}}$ $=\frac{\text{2m}^2-\text{m}-\text{m}+1}{\text{2m}^2-\text{2m}+1}$ $=\frac{\text{2m}^2-\text{2m}+1}{\text{2m}^2-\text{2m}+1}$ $=1$ $\therefore\tan\text{(A-B)}=1=\Big(\tan\frac{\pi}{4}\Big)$ $\Big[\because\tan\frac{\pi}{4}=1\Big]$ $\Rightarrow\tan\text{(A-B)}=1=\Big(\tan\frac{\pi}{4}\Big)$ $\Rightarrow\text{A-B}=\frac{\pi}{4}$

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