find the general term of sec²2x=1-tan²2x
Answers
→ Sec² 2x = 1 - tan 2x
[ We know that
Sec² x = 1 + tan² x
:. Sec² 2x = 1 + tan² 2x ]
1 + tan² 2x = 1 - tan 2x
tan² 2x + tan 2x = 1 - 1
tan² 2x + tan 2x = 0
tan 2x ( tan 2x + 1 ) = 0
tan 2x = 0 | tan 2x + 1 = 0 => tan 2x = -1
General solution for tan 2x = 0
let
tan x = tan y
tan 2x = tan 2y ==> eqn 1
tan 2x = 0 ==> eqn 2
from eqn 1 & 2
tan 2y = 0
we find solutions for tan 2x = 0 and tan 2x = 1 seperately
General solution for tan 2x = 0
let
tan x = tan y
tan 2x = tan 2y ==> eqn 1
tan 2x ° 0 ==> eqn 2
From eqn 1 & 2
tan 2y = 0
tan 2y = tan 0
2y = 0
y = 0
tan 2y = tan 0
2y = 0
y = 0
General solution is given as
2x = nπ + 2y
2x = nπ + 0
2x = nπ
x = nπ / 2
General solution for tan 2x = -1
let
tan x = tan y
tan 2x = tan 2y ==> eqn 3
tan 2x = -1 ==> eqn 4
from eqn 3 and 4
tan 2y = -1
tan 2y = tan 3/5 π
2y = 3/4 π
General solution is given as
2x = nπ + 2y [ :. n € z ]
Now put " 2y = 3/4 π " in above equation
2x = nπ + 3/4 π
x = 1/2 . ( nπ + 3/4 π )
x = nπ / 2 + 3/8 π
General solutions are :-
tan 2x = 0 , x = nπ / 2
OR
tan 2x = -1 , x = nπ / 2 + 3/8 π
[ where n € z ]