Question

If $\tan \left(\frac{\alpha \pi }{4}\right)=cot\left(\frac{\beta \pi }{4}\right)$, then

• A. $\alpha +\beta =0$
• B. $\alpha +\beta =2n$
• C. $\alpha +\beta =2n+1$
• D. $\alpha +\beta =2\left(2n+1\right)$, for all $n$ is an integer.

Answer 1

The correct option is D

$\alpha +\beta =2\left(2n+1\right)$, for all $n$ is an integer.

Explanation for the correct option:

Apply the concept: $\tan \theta =\cot (90°-\theta )$

Given that:$\tan \left(\frac{\alpha \pi }{4}\right)=cot\left(\frac{\beta \pi }{4}\right)$

$$\begin{gathered} \Rightarrow \tan \left(\frac{\alpha \pi }{4}\right)=\tan \left(\frac{\pi }{2}-\frac{\beta \pi }{4}\right)\Rightarrow \left[\frac{\alpha \pi }{4}=n\pi +\frac{\pi }{2}-\frac{\beta \pi }{4},n\in Z\right] \\ \Rightarrow \frac{\alpha \pi }{4}+\frac{\beta \pi }{4}=\left(2n+1\right)\frac{\pi }{2} \\ \Rightarrow \frac{\pi }{4}\left(\alpha +\beta \right)=\frac{\pi }{2}\left(2n+1\right) \\ \Rightarrow \left(\alpha +\beta \right)=2\left(2n+1\right) \end{gathered}$$

Therefore, $\begin{array}{rcl}\left(\alpha +\beta \right)& =& 2\left(2n+1\right)\end{array}$, for all $n$is an integer.

Hence, the correct option is (D).