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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 Angles1 Mark
Question
If $\tan\alpha=\frac{1}{1+2^{-\text{x}}}$ and $\tan\beta=\frac{1}{1+2^{\text{x+1}}},$ then write the value of $\alpha+\beta$ lying in the interval $\Big(0,\frac\pi2\Big).$
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Answer

We have, $\tan\alpha=\frac{1}{1+2^{-\text{x}}}$ and $\tan\beta=\frac{1}{1+2^{\text{x+1}}}$ Now, $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\times\tan\beta}$ $=\frac{\frac{1}{1+2^{-\text{x}}}+\frac{1}{1+2^{\text{x+1}}}}{1-\frac{1}{1+2^{-\text{x}}}\times\frac{1}{1+2^{\text{x+1}}}}$ $=\frac{\frac{1+2^{\text{x+1}}+1+2^{-\text{x}}}{(1+2^{-\text{x}})(1+2^{\text{x+1}})}}{1-\frac{1}{(1+2^{-\text{x}})(1+2^{\text{x+1}})}}$ $=\frac{\frac{2+2^{\text{x}+1}+2^{-\text{x}}}{1+2^{\text{x}+1}+2^{-\text{x}}+2}}{1-\frac{1}{1+2^{\text{x}+1}+2^{-\text{x}}+2}}$ $=\frac{\frac{2+2^{\text{x}+1}+2^{-\text{x}}}{3+2^{\text{x}+1}+2^{-\text{x}}}}{1-\frac{1}{3+2^{\text{x}+1}+2^{-\text{x}}}}$ $=\frac{\frac{2+2^{\text{x}+1}+2^{-\text{x}}}{3+2^{\text{x}+1}+2^{-\text{x}}}}{\frac{3+2^{\text{x}+1}+2^{-\text{x}}-1}{3+2^{\text{x}+1}+2^{-\text{x}}}}$ $=\frac{2+2^{\text{x+1}}+2^{-\text{x}}}{2+2^{\text{x+1}}+2^{-\text{x}}}$ $=1$ $\Rightarrow\tan(\alpha+\beta)=1=\tan\Big(\frac{\pi}{4}\Big)$ $\Rightarrow\tan(\alpha+\beta)=\tan\Big(\frac\pi4\Big)$ $\Rightarrow\alpha+\beta=\frac\pi4$

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