# pl. provide solutions for the problems in attached sheet

Kaustubh Nayyar
27 Points
7 years ago
hey I got the solution of the first problem
(1) first, open tan both sides by changing it in terms tan pi/4, tan x/2 and tan y/2
the expresion becomes                         ( / means divided by)
(1+tan y/2) / (1 – tan y/2) = [ ( 1+tan x/2 ) / ( 1 – tan x/2) ]3
(2) now change tan x/2 and tan y/ 2 in terms of sin and cos
the expresion becomes
(cos y/2 + sin y/2) / (cos y/2 – sin y/2) =
[ ( cos x/2 + sin x/2 ) / ( cos x/2 – sin x/2) ]3
(3)  square the  L.H.S. and open   square the R.H.S and open the square but not the cubic power , let the cubic power as it is
L.H.S. = ( 1 + 2sin y/2 cos y/2) / ( 1 –   2sin y/2 cos y/2)
= ( 1 + sin y ) / ( 1 – sin y )
R.H.S   =  [ ( 1 + sin x ) / ( 1 – sin x ) ]3
(4)     now apply componendo and dividendo both sides LHS & RHS
( componendo divindendo means that if a/b = c/d then
(a+b) / (a – b) = (c+d) / (c – d)  )
we get      2 / 2sin y =
( 1 + sin x )3 + ( 1 – sin x )] / [ ( 1 + sin x )3 –  ( 1 – sin x )]
(5)   now open the cubes
you will get the required answer