Question

If `tan(pi/2sintheta) = cot(pi/2costheta)` then sum of all possible values of `sintheta + costheta` equal to?

Answer 1

`0`

Explanation:

If `\tan A = tan B`, then `A-B = n\pi`, where `n \in \mathbbZ`. Since you can write the given equation as `\tan (\frac{\pi}2 \sin \theta ) = \tan (\frac{\pi}2 - \frac{\pi}2 \cos \theta)`, we have that `\frac{\pi}2 \sin \theta - \pi/2 + \pi/2 \cos \theta = n \pi`, or `\sin \theta + \cos \theta = 1/2 + n`, where `n \in \mathbbZ`.

Of course, `\sin \theta + \cos \theta` can only lie between `-\sqrt2` and `\sqrt2`, so the only possible values are at `n = 0` and `n = -1`, and they are `1/2` and `-1/2`. That's it!