Question

the value of sec pi by 7,sec 3 pi by 7 ,sec 5 pi by 7​

Answer 1

Given :  sec(π/7).sec(3π/7).Sec(5π/7)

To Find :  Value

Solution:

sec(π/7).sec(3π/7).Sec(5π/7)

use secx = 1/cosx

= 1/ cos(π/7)cos(3π/7).cos(5π/7)

multiply and divide by 2sin(π/7)

= 2sin(π/7)/2sin(π/7)cos(π/7)cos(3π/7).cos(5π/7)

use 2sinxcosx = sin2x

= 2sin(π/7)/ sin(2π/7)cos(3π/7).cos(5π/7)

Cosx  = -Cos(π - x)

=> cos(5π/7) = - cos(2π/7)

= 2sin(π/7)/ sin(2π/7)cos(3π/7).(-cos(2π/7))

= - 2sin(π/7)/ sin(2π/7)cos(2π/7)cos(3π/7)

= - 4sin(π/7)/ 2sin(2π/7)cos(2π/7)cos(3π/7)

=  - 4sin(π/7)/ sin(4π/7)cos(3π/7)

cos(3π/7)  = - Cos(4π/7)

=  - 4sin(π/7)/ sin(4π/7)(- Cos(4π/7))

= 4sin(π/7)/ sin(4π/7) Cos(4π/7)

= 8 sin(π/7)/ 2sin(4π/7) Cos(4π/7)

= 8 sin(π/7)/ sin(8π/7)

Sinx = - sin(π + x)

sin(8π/7) = -  sin(π/7)

= 8 sin(π/7)/ -sin(π/7)

= - 8

sec(π/7).sec(3π/7).Sec(5π/7) = - 8

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Answer 2

$\huge\purple{\fbox{\sf{Dear \: User}}}$

$\mathcal\pink{Due \: to \: No \: symbol \: pi \: Iam \: Writing \: as 'T'}$

$sec(T/7).sec(3T/7).Sec(5T/7) \\ use \: secx = 1/cosx \\ 1/ cos(T/7)cos(3T/7).cos(5T/7) \\ multiply \: and \: divide \: by \: 2sin(T/7) \\ = 2sin(T/7)÷2sin(T/7)cos(T/7)cos(3T/7)×cos(5T/7) \\ use \: 2sinxcosx = sin2x \\ 2sin(T/7)÷ sin(2T/7)cos(3T/7).cos(5T/7) \\ Cosx = -Cos(T- x) \\ ==>cos(5T/7) = - cos(2T/7) \\ =2sin(T/7)÷ sin(2T/7)cos(3T/7)(cos(2T/7) \\ = 2sin(T/7) ÷ sin(2T/7)cos(2T/7)cos(3T/7) \\ = \: - 4sin(T/7)÷2sin(21/7)cos(21/7)cos(3T/7) \\ = - 4sin(T/7)÷sin(4T/7)cos(3T/7) \\ cos(3T/7) = \: -Cos(4T/7) \\ = - 4sin(T/7)÷sin(4T/7)- Cos(4T/7) \\ 4sin(T/7)÷sin(4T/7) Cos(4T/7) \\ = 8 sin(T/7)÷ 2sin(4T/7) Cos(4T/7) \\ -8 sin(T/7)÷ sin(8T/7) \\ Sinx= \: - sin(T + x) \\ sin(8T/7) = \: - sin(T/7) \\ 8 sin(T/7)/-sin(T/7)\\ = - 8$