यदि `theta = pi/8` है, तो `(2 cos theta + 1)^(10)` , `(2cos 2 theta-1)^(10) (2 cos theta - 1)^(10)` , `(2 cos 4 theta- 1)^(10)` का मान है ?

Options

A. 0
B. 1
C. 2
D. 4

Explanation

यदि `theta = (pi)/(8)`,
`(2 cos theta + 1)^(10) (2 cos 2theta - 1)^(10)`
`(2 cos theta - 1)^(10) (2 cos 4 theta - 1)^(10)`
`rArr (2 cos theta + 1)^(10) (2 cos theta - 1)^(10)`
`(2 cos 2theta - 1)^(10) (2 cos 4 theta - 1)^(10)`
`rArr (4 cos^(2) theta - 1)^(10) (2 cos 2 theta - 1)^(10)`
`(2 cos 4 theta - 1)^(10)`
`rArr [ 2(2 cos^(2) theta]^(10) (2 cos 2 theta - 1)^(10)`
`(2 cos 4 theta - 1)^(10)`
`rArr [2 (1+cos2 theta) - 1]^(10) (2 cos 2 theta - 1)^(10)`
`(2 cos 4 theta - 1)^(10)`
`rArr (2 cos 2 theta + 1)^(10) (2 cos 2 theta - 1)^(10)`
`(2 cos 4 theta - 1)^(10)`
`rArr [ 2 cos ' (pi)/(4) + 1]^(10) (2 cos '(pi)/(4) - 1)^(10)`
`(2 cos '(pi)/(2) - 1), [ :' theta = (pi)/(8)]`
`rArr [(2 xx 1/(sqrt(2)) + 1)^(10)(2.(1)/(sqrt(2)) - 1)^(10) (0-1)^(10)`
`=(sqrt(2) + 1)^(10) (sqrt(2) - 1)^(10) xx 1`
`= (sqrt(2) + 1)^(10) (sqrt(2) -1)^(10) xx 1`
`= ((sqrt(2))^(2) - (1)^(2))^(10) xx 1`
`= (2-1)^(10)`
`= 1`