1. Tardigrade
  2. Question
  3. Mathematics
  4. Statement-I: If displaystyle∑ i =12 n displaystyle sin -1 x 1= n π, n ∈ N. Then displaystyle∑ i =1 n x i = displaystyle∑ i =1 n x 12= displaystyle∑ i =1 n x 13 Because Statement-II: -(π/2) ≤ sin -1 x ≤ (π/2), ∀ x ∈[-1,1]

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Q. Statement-I : If $\displaystyle\sum_{ i =1}^{2 n } \displaystyle\sin ^{-1} x _1= n \pi, n \in N$. Then $\displaystyle\sum_{ i =1}^{ n } x _{ i }=\displaystyle\sum_{ i =1}^{ n } x _1^2=\displaystyle\sum_{ i =1}^{ n } x _1^3$ Because
Statement-II : $-\frac{\pi}{2} \leq \sin ^{-1} x \leq \frac{\pi}{2}, \forall x \in[-1,1]$

Inverse Trigonometric Functions

Solution:

Correct answer is (a) Statement-I is true, Statement-II is true; Statement-II is correct explanation for Statement-I.