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If \[{{\sin }^{-1}}x+{{\sin }^{-1}}y+{{\sin }^{-1}}z=\dfrac{3\pi }{2}\] the value of \[{{x}^{100}}+{{y}^{100}}+z{}^{100}-\dfrac{9}{{{x}^{101}}+{{y}^{101}}+{{z}^{101}}}\] is ?
A. 0
B. 1
C. 2
D. 13


Answer
VerifiedVerified
162k+ views
Hint:
We have to know the range and values of inverse sine functions to solve the given problem. We know the domain of function varies from ( -1, 1) and ranges from \[\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)\]. To solve the question, we divide the RHS and on the LHS we have a sum of three inverse functions that take the maximum value of \[ \dfrac{\pi }{2}\]. By putting the value equal to \[\dfrac{\pi }{2}\], we get the value of x, y and z. Then we solve the equations to get our answer.

Complete Step- by- step Solution
Given \[{{\sin }^{-1}}x+{{\sin }^{-1}}y+{{\sin }^{-1}}z=\dfrac{3\pi }{2}\]
This will happen only when \[{{\sin }^{-1}}x={{\sin }^{-1}}y={{\sin }^{-1}}z=\dfrac{\pi }{2}\]
We have to find the value of \[{{x}^{100}}+{{y}^{100}}+z{}^{100}-\dfrac{9}{{{x}^{101}}+{{y}^{101}}+{{z}^{101}}}\]
We know \[{{\sin }^{-1}}x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)\]
Sum of three inverse of sine is \[3\times \dfrac{\pi }{2}\]
That is every sine inverse function is equal to \[\dfrac{\pi }{2}\]
That is $x = \dfrac{3\pi }{2} \sin \dfrac{\pi }{2}, y = \sin \dfrac{\pi }{2}, z = \sin \dfrac{\pi }{2}$
This means $\sin \dfrac{\pi }{2}=1$
That is $x = y = z = 1$
 Hence the value of ${{x}^{100}}+{{y}^{100}}+z{}^{100}-\dfrac{9}{{{x}^{101}}+{{y}^{101}}+{{z}^{101}}} = 1 + 1 + 1-\dfrac{9}{1+1+1}$
By solving the above equation, we get $3-\dfrac{9}{3}= 0$
Hence the value of ${{x}^{100}}+{{y}^{100}}+z{}^{100}-\dfrac{9}{{{x}^{101}}+{{y}^{101}}+{{z}^{101}}} = 0$

Thus, Option (A) is correct.

Note The domain of function of sin inverse is ( -1,1) and range is $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$. That means the maximum value that the inverse sine function can take is $\dfrac{\pi }{2}$. If we observe that the given problem on the RHS value is $\dfrac{3\pi }{2}$ and on the LHS we have a sum of three inverse functions. the sum can be achieved a value of $\dfrac{3\pi }{2}$, if and only if each inverse sine function takes their maximum value. That is the main thing we take care while solving these type of questions.