Question

The number of triplets \[\left( {x,y,z} \right)\] satisfying ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\cos ^{ - 1}}z = 2\pi $ is
A. 1
B. 0
C. 2
D. $\infty $

Answer 1

Hint: We will first write the range of all the given terms. Then, we will find the combination of values for the given condition holds. The number of values possible of triplet \[\left( {x,y,z} \right)\] is the required answer. A required triplet will be the possible values of $x,y,z$ which will satisfy the given condition. The order of $x,y,z$ is important in writing the triplet.

Complete step by step Answer:

We know that the range of the ${\sin ^{ - 1}}\theta $ is $\left[ { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right]$
That is $ - \dfrac{\pi }{2} \leqslant {\sin ^{ - 1}}x \leqslant \dfrac{\pi }{2}$
Similarly, $ - \dfrac{\pi }{2} \leqslant {\sin ^{ - 1}}y \leqslant \dfrac{\pi }{2}$
The range of ${\cos ^{ - 1}}\theta $ is $\left[ {0,\pi } \right]$
Then, $0 \leqslant {\cos ^{ - 1}}z \leqslant \pi $
We want to make the sum ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\cos ^{ - 1}}z = 2\pi $
Thus, we will have to extreme values of all the terms to get the required sum.
That is,
${\sin ^{ - 1}}x = \dfrac{\pi }{2}$
${\sin ^{ - 1}}y = \dfrac{\pi }{2}$
${\cos ^{ - 1}}z = \pi $
Now, we write the corresponding values of $x,y,z$
When ${\sin ^{ - 1}}x = \dfrac{\pi }{2}$, we know that $\sin \dfrac{\pi }{2} = x$
Then, $x = 1$
Similarly, $y = 1$
Now,
 $
  {\cos ^{ - 1}}z = \pi \\
   \Rightarrow z = \cos \pi \\
   \Rightarrow z = - 1 \\
$
Therefore, the possible triplet is $\left( {1,1, - 1} \right)$ and there is only one such triplet possible.
Hence, option A is correct.

Note: One must remember the range for given trigonometric functions in order to avoid errors. The inverse trigonometry functions reverse the operations that the sine, cosine tangent, secant, cosecant, and cotangent perform. Here, many students make mistakes by not considering the range of the inverse trigonometric functions.