Question

Solve & find the general solution of:

Answer 1

\[\tan^2\frac{ \theta }{ 2 }-1=0\]

Answer 2

im sorry i can't see anything??

Answer 3

i didn't get it u cant see <_<

Answer 4

ii meant i don't see any equation

Answer 5

well the question itself is an equation

Answer 6

i think solving for theta gives Theta=2*arctan(1)

Answer 7

nvm he/shes got it

Answer 8

\[\tan \frac{ \theta }{ 2 }=\pm \sqrt{\frac{ 1-\cos \theta }{ 1+\cos \theta }}\]

Answer 9

is that a new question? or was it the first one?

Answer 10

why to convert it to cos, keep it in tan
tan (x/2) = +1 or -1

Answer 11

what how?

Answer 12

tan^2 (x/2) -1 =0
tan^2 (x/2) = 1
tan (x/2) =1 or tan (x/2) = -1

Answer 13

x/2 = 45 (or pi/4) or x/2 = -(pi/4)
x=...? or .... ?

Answer 14

tan x/2 =1 how to do it?

Answer 15

tan value = 1 at angles, 45, 45+180,45+2*180,... 45 +n*180

Answer 16

so, tan (x/2) = 1
implies (x/2) = pi/4 +n * pi

Answer 17

x= 2*pi/4 + 2*n*pi

Answer 18

x= pi/2 +2n *pi
same for tan (x/2) =-1

Answer 19

\[{n \pi \pm \frac{ \pi }{ 2 }}\]

Answer 20

this is the answer

Answer 21

and we got the same ?
its like
tan (x/2) = 1
implies (x/2) = pi/4
x= 2*pi/4 = pi/2
and since tan function is periodic in pi,
x = pi/2 +n*pi
for
tan (x/2) = -1
x/2 = -pi/4
x = -pi/2 +n*pi

Answer 22

got that ^ ?

Answer 23

ok