Question

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

Answer 1

Hint: we have to find the value of x, y and z from the given equation using the values of trigonometric standard angles. The domain and range of sine inverse function is (-1,1) and $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$ respectively. This will be used in finding the values of x, y and z. Then, we have to put those values in the equation whose answer we have to find and by solving step by step we will get our answer.

Complete step by step answer:
Given:
${\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \dfrac{{3\pi }}{2}$
We can also rewrite this equation as
 ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z = \dfrac{\pi }{2} + \dfrac{\pi }{2} + \dfrac{\pi }{2}$
So, this means the maximum value in the range of sine inverse x is $\dfrac{\pi }{2}$ .
Sum of three inverse of sine is $3 \times \dfrac{\pi }{2}$
This states that every sine inverse function is equal to $\dfrac{\pi }{2}$. Then,
${\sin ^{ - 1}}x = \dfrac{\pi }{2}$, ${\sin ^{ - 1}}y = \dfrac{\pi }{2}$ and ${\sin ^{ - 1}}z = \dfrac{\pi }{2}$
So, $x = \sin \dfrac{\pi }{2}$, $y = \sin \dfrac{\pi }{2}$and $z = \sin \dfrac{\pi }{2}$
We know that the value of $\sin \dfrac{\pi }{2}$is 1. So,
X = 1, y = 1 and z = 1.
Now we will put value of x, y and z in the equation ${x^9} + {y^9} + {z^9} - \dfrac{1}{{{x^9}{y^9}{z^9}}}$and we will get,
${x^9} + {y^9} + {z^9} - \dfrac{1}{{{x^9}{y^9}{z^9}}} = {1^9} + {1^9} + {1^9} - \dfrac{1}{{{1^9} \times {1^9} \times {1^9}}}$
${x^9} + {y^9} + {z^9} - \dfrac{1}{{{x^9}{y^9}{z^9}}} = 1 + 1 + 1 - \dfrac{1}{{1 \times 1 \times 1}}$
${x^9} + {y^9} + {z^9} - \dfrac{1}{{{x^9}{y^9}{z^9}}} = 3 - \dfrac{1}{1}$
${x^9} + {y^9} + {z^9} - \dfrac{1}{{{x^9}{y^9}{z^9}}} = 3 - 1$
${x^9} + {y^9} + {z^9} - \dfrac{1}{{{x^9}{y^9}{z^9}}} = 2$

So, the correct answer is “Option C”.

Note: The range of sine inverse function is $\left( { - \dfrac{\pi }{2},\dfrac{\pi }{2}} \right)$. So, the maximum value that sine inverse function can take is $\dfrac{\pi }{2}$. Because of this we have split $\dfrac{{3\pi }}{2}$into three $\dfrac{\pi }{2}$. Sum of ${\sin ^{ - 1}}x + {\sin ^{ - 1}}y + {\sin ^{ - 1}}z$ can be $\dfrac{{3\pi }}{2}$only if value of each function is equal or less than $\dfrac{\pi }{2}$. This thing is very important to solve the question. And same is with other trigonometric functions.