MCQ
Choose the correct answer.
If
$\tan\alpha=\frac{\text{m}}{\text{m}+1},\tan\beta=\frac{1}{2\text{m}+1},$ then $\alpha+\beta$ is equal to:- A$\frac{\pi}{2}$
- B$\frac{\pi}{3}$
- C$\frac{\pi}{6}$
- D$\frac{\pi}{4}$
If
$\tan\alpha=\frac{\text{m}}{\text{m}+1},\tan\beta=\frac{1}{2\text{m}+1},$ then $\alpha+\beta$ is equal to:Solution:
Given that, $\tan\alpha=\frac{\text{m}}{\text{m}+1}$ and $\tan\beta=\frac{1}{2\text{m}+1}$
Now, $\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\cdot\tan\beta}=\frac{\frac{\text{m}}{\text{m}+1}+\frac{1}{2\text{m}+1}}{1-\Big(\frac{\text{m}}{\text{m}+1}\Big)\Big(\frac{1}{2\text{m}+1}\Big)}$
$=\frac{\text{m}(2\text{m}+1)+\text{m}+1}{(\text{m}+1)(2\text{m}+1)-\text{m}}=\frac{2\text{m}^2+2\text{m}+1}{2\text{m}^2+3\text{m}+1-\text{m}}=1$
$\therefore\alpha+\beta=\frac{\pi}{4}$
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