MHT CET · Maths · Inverse Trigonometric Functions
If \(\cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi\), then the value of \(x^{2025}+x^{2026}+x^{2027}\) is
- A \(-1\)
- B \(0\)
- C \(1\)
- D \(3\)
Answer & Solution
Correct Answer
(A) \(-1\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \cos ^{-1} x+\cos ^{-1} y+\cos ^{-1} z=3 \pi \\ & \text { Since } 0 \leq \cos ^{-1} x \leq \pi \\ & \\ & 0 \leq \cos ^{-1} y \leq \pi \text { and } 0 \leq \cos ^{-1} z \leq \pi \\ & \\ & \text { Here, } \cos ^{-1} x=\cos ^{-1} y=\cos ^{-1} z=\pi \\ & \Rightarrow x=y=z=\cos \pi=-1 \\ & \therefore \quad x^{2005}+x^{2026}+x^{2027} \\ & =(-1)^{2025}+(-1)^{2026}+(-1)^{2027} \\ & =-1+1-1 \\ & =-1\end{aligned}\)
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