MHT CET · Maths · Inverse Trigonometric Functions
If \(\sin ^{-1} x+\sin ^{-1} \mathrm{y}+\sin ^{-1} \mathrm{z}=\frac{3 \pi}{2}, \quad\) then \(x^{100}+\mathrm{y}^{100}+\mathrm{z}^{100}=\)
- A \(3\)
- B \(4\)
- C \(2\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(3\)
Step-by-step Solution
Detailed explanation
Given
\(\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2} \Rightarrow \sin ^{-1} x=\sin ^{-1}\) \( y=\sin ^{-1} z=\frac{\pi}{2} \)
\( \therefore x=y=z=1 \)
\( \therefore x^{100}+y^{100}+z^{100}=1+1+1=3\)
\(\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2} \Rightarrow \sin ^{-1} x=\sin ^{-1}\) \( y=\sin ^{-1} z=\frac{\pi}{2} \)
\( \therefore x=y=z=1 \)
\( \therefore x^{100}+y^{100}+z^{100}=1+1+1=3\)
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