JEE Class main Answered

Let cot^-1 ( - cot 5x ) = tan^-1 ( cot x ) = α

from LHS  ,  cot^-1 ( - cot 5x ) =  α

cot( α ) = - cot(5x) ........................... (1)

Similarly from RHS

 tan^-1 ( cot x ) = α

tan ( α ) = cot x  ..........................(2)

From eqn.(1) and (2) , we get

cot ( α ) tan (α ) = - cot(5x) cot(x) = 1

hence we get, cos(5x)cos(x) + sin(5x) sinx = 0 

cos( 5x -x ) = cos(4x) = 0

cos(4x) = cos[ (2n+1) (π / 2 ) ] , where n = 0, 1,  2 ........ etc

4x = (2n+1) (π / 2 )

x = (2n+1) (π / 8 )

if n = 0 , then x = (π / 8 )

if n =1 , then x = (3 π ) / 8