Q197 / 270 · NDA 2024 (Phase 1) · Mathematics

If $f(\theta)=\frac{1}{1+\tan \theta}$ and $\alpha+\beta=\frac{5 \pi}{4}$, then what is the value of $f(\alpha) f(\beta)$ ?

  • A
    $-\frac{1}{2}$
  • Correct answer
    B
    $\frac{1}{2}$
  • C
    1
  • D
    2

Solution

$\alpha+\beta=\frac{5 \pi}{4}$ $\begin{aligned} & \tan (\alpha+\beta)=\tan 5 \frac{\pi}{4} \\ & \Rightarrow \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \times \tan \beta}=\tan \left(\pi+\frac{\pi}{4}\right)=\tan \frac{\pi}{4} \\ & \Rightarrow \frac{\tan \alpha+\tan \beta}{1-\tan \alpha \times \tan \beta}=1 \\ & \Rightarrow \tan \alpha+\tan \beta=1-\tan \alpha \times \tan \beta \\ & \Rightarrow \tan \alpha+\tan \beta+\tan \alpha \cdot \tan \beta=1 \\ & \because f(\theta)=\frac{1}{1+\tan \theta} \\ & f(\alpha) \cdot f(\beta)=\frac{1}{1+\tan \alpha+\tan \beta+\tan \alpha \cdot \tan \beta}=\frac{1}{1+1}=\frac{1}{2} .\end{aligned}$

Hence, the correct answer is option: 'B'.

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