Question

The least positive integer n for which $\left( \frac{1 + i}{1 - i} \right)^{n}$= $\frac { 2 } { \pi }$ $\left( \sec^{- 1}\frac{1}{x} + \sin^{- 1}x \right)$

(where x¹ 0; –1 £ x £ 1) is

• A. 2
• B. 4
• C. 6
• D. 8

Answer 1

Answer: (B)

4

Answer 2

Sol. In –1 £ x £ 1, sec^–1 $\left( \frac{1}{x} \right)$= cos^–1 x

\ sec^–1 $\left( \frac{1}{x} \right)$+ sin^–1 x = cos^–1 x + sin^–1 x = $\frac{\pi}{2}$

\ $\left( \frac{1 + i}{1 - i} \right)^{n}$= 1 Ū $\left\lbrack \frac{(1 + i)^{2}}{2} \right\rbrack^{n}$= 1 Ū iⁿ = 1

Hence least positive integral value of n = 4.