The number of triplets satisfying $sin^{-1}x + cos^{-1} y + tan^{-1} z = 2 \pi , $ is
Answer & explanation
Correct answer: option 1
We know that
$sin^{-1} x ≤ \frac{\pi}{2},\cos^{-1} y ≤ \pi, \tan^{-1} z ≤ \frac{\pi}{2}$
so $\sin^{-1}x+\cos^{-1}y+\tan^{-1} z<\frac{\pi}{2}$
No triplet exists.