If 5 sin x cos y = 1, 4 tan x = tan y, then

Options

A. `x=(m+n)(pi)/(2)+(pi)/(4)+(-1)^(m)(1)/(2)sin^(-1)(-(3)/(5)),m, n in Z`
B. `y=(n-m)(pi)/(2)+(pi)/(4)+(-1)^(m+1)(1)/(2)sin^(-1)(-(3)/(5)),m,n in Z`
C. `x=(m+n)(pi)/(2)+(pi)/(4)+(-1)^(m)(1)/(2)sin^(-1)((3)/(5)),m,n in Z`
D. `y=(n-m)(pi)/(2)+(pi)/(4)+(-1)^(m+1)(1)/(2)sin^(-1)((3)/(5)),m,n in Z`

Answer

A, B

Explanation

We have 5 sin x cos y = 1 and 4 tan x = tan y
`rArr 5 sin x cos y=1` and 4 sin x cos y = sin y cos x
`rArr sin x cos y=(1)/(5)` and `cos x sin y = (4)/(5)`
`rArr sin x cos y + cos sin y =1`
and `sin x cos y-cos x sin y=-()3)/(5)`
`rArr sin(x+y)=sin.(pi)/(2)` and `sin(x-y)=sin{sin^(-1)(-(3)/(5))}`
`rArr x+y =(2n+1)(pi)/(2), n in Z`
and `x-y = m pi (-1)^(m)sin^(_1)(-(3)/(5)), m in Z`
`rArr x = (m+n)(pi)/(2)+(pi)/(4)+(-1)^(m)(1)/(2)sin^(-1)(-(3)/(5)), m, n in Z`
and `y=(n-m)(pi)/(2)+(pi)/(4)+(-1)^(m+1)(1)/(2)sin^(-1)(-(3)/(5)), m, n in Z`