Solve the equation `2 tan^(-1) (cos x) = tan^(-1) (2 cosec x)`

Answer

`a rarr p, r, s; b rarr p`

Explanation

Given `2(a^(2) -b^(2)) = c^(2)`
`rArr 2 (sin^(2) X - sin^(2)Y) = sin^(2)Z`
`rArr 2 sin (X + Y) sin (X - Y) = sin^(2) Z`
`rArr 2 sin (pi - Z) sin (X -Y) = sin^(2) Z`
`rArr sin (X - Y) = (sin Z)/(2)`...(i)
`:. lamda = (sin (X -Y))/(sin Z) = (1)/(2)`
Now `cos (n pi lamda) = 0`
`rArr cos ((n pi)/(2)) = 0`
`:. n = 1, 3, 5`

`1 + cos 2X - 2 cos 2Y = 2 sin X sin Y`
`2 cos^(2) X - 2 cos 2Y = 2 sin X sin Y`
`1- sin^(2) X - 1 + 2 sin^(2) Y = sin X sin Y`
`sin^(2) X + sin X sin Y = 2 sin^(2) Y`
`sin X (sin X + sin Y) = 2 sin^(2) Y`
`rArr a(a +b) = 2b^(2)`
`rArr a^(2) + ab - 2b^(2) = 0`
`rArr ((a)/(b))^(2) + (a)/(b) - 2 = 0`
`rArr (a)/(b) = -2, 1`
`rArr (a)/(b) = 1`
Note : Solutions of the remaining parts are given in their respecitive chapters.