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

Options

A. `f_(2) ((pi)/(16))=1`
B. `f_(3) ((pi)/(32))=1`
C. `f_(4) ((pi)/(64))=1`
D. `f_(5) ((pi)/(128))=1`

Answer

A, B, C, D

Explanation

NOTE Multiplicative loop is very important approach in IIT Mathematics
`(tan.(theta)/(2))(1+sectheta)=(sintheta//2)/(sintheta//2)[1+(1)/(costheta)]`
`=((sintheta//2)2cos^(2)theta//2)/((costheta//2)costheta)`
`=((2sintheta//2)cos theta//2)/(costheta)=(sintheta)/(costheta)=tan theta`
`therefore" " f_(n)(theta) = (tantheta//2)(1+sectheta)`
`(1+sec 2 theta)(1+sec2^(2)theta)...(1+sec2^(n)theta)`
`=(tan theta)(1+sec 2 theta)(1+sec2^(2) theta)...(1+sec2^(n)theta)`
`=tan2 theta.(1 + sec2^(2)theta)...(1+sec2^(n)theta)`
`=tan (2^(n)theta)`
Now, `f_(2)((pi)/(16))=tan(2^(2).(pi)/(16))=tan((pi)/(4))=1`
Therefore, (a) is the answer.
`f_(3)((pi)/(32))=tan(2^(3).(pi)/(32))=tan((pi)/(4))=1`
Therefore, (b) is the answer.
`f_(4)((pi)/(64))=tan(2^(4).(pi)/(64))=tan((pi)/(4))=1`
Therefore, (c) is the answer.
`f_(5)((pi)/(128))=tan(2^(5).(pi)/(128))=tan((pi)/(4))=1`
Therefore, (d) is the answer.