For a positive integer `n ,` let `f_n(theta)=(tantheta//2)(1+sectheta)(1+sec2theta)(1+sec4theta).......... (1+sec2^ntheta)`. Then (a)`f_2(pi/(16))=1` (b) `f_3(pi/(32))=1` (c)`f_4(pi/(64))=1` (d) `f_5(pi/(128))=1`

Options

A. `f_(2)((pi)/(64))=1`
B. `f_(3)((pi)/(32))=1`
C. `f_(4)((pi)/(32))=1`
D. `f_(5)((pi)/(128))=-1`

Answer

D

Explanation

`f_(n)(theta)`
`=(tan(theta)/(2))(1+sectheta)(1+sec2 theta)(1+sec 4 theta)...(1+sec 2^(n)theta)`
`=(tan""(theta)/(2))((1+costheta)/(costheta))(1+sec2 theta)(1+sec 4theta)...(1+sec2^(n)theta)`
`=(tan""(theta)/(2))xx(2 cos^(2)theta)/(cos theta)(1+sec2theta)(1+sec4theta)...(1+sec2^(n)theta)`
`=tantheta(1+sec2theta)(1+sec 4theta)...(1+sec2^(n)theta)`
`=tan2theta(1+sec4theta)...(1+sec2^(n)theta)=tan2^(n)theta`
`thereforef_(2)((pi)/(16))=tan(2^(4)xx(pi)/(16))=1,f_(3)((pi)/(32))=tan(2^(3)xx(pi)/(32))=1`
`f_(4)""(pi)/(64)=tan(2^(4)xx(pi)/(64))=1,f_(5)((pi)/(128))=tan(2^(5)xx(pi)/(128))=1`
Thus, all the options are true.