how to prove tan^2x+cot^2x=2sec^x-1+cosec^2x-1=2

by the definition of tan x , tan x =sin x/cos
x

therefore, tan^2 x= sin^2 x/cos^2
x

The definition of cot x states that cot x=cos x/sin
x

square both sides, you
have:

cot^2 x= cos^2 x/sin^2
x

substitue these values in the
function

tan^2 x+ cot^2 x= sin^2 x/cos^2 x + cos^2 x/sin^2
x

simplify into one
fraction

1.(sin^4 x +cos ^4 x)/(cos^2 x*sin^2
x)

using the binomial
theorem

2. sin^4 x + cos ^4 x= (sin^2 x +cos^2 x)^2 -
2*sin^2 x * cos^2 x

sin^2 x + cos^2 x=
1

function 2 becomes

sin^4 x+
cos^4 x= 1-2sin^2x*cos^2 x

3. 2sin^2 x*cos^2 x= 2(sin x*cos
x)^2

by the double-angle
formula

sin 2x= 2 sin x cos
x

sin2x/2=sin x cos x

function
number 3 becomes (sin 2x/2)^2

=sin^ 2
2x/4

put that back into function
3

2* sin ^2 2x/4=sin^2
2x/2

put that back into function
2

sin^4 x+ cos^4 x= 1-2sin^2x*cos^2
x

= 1- sin^2 2x/2

the
denominator could also use the double formula

so function
one becomes

(1-sin^2 2x/2 )/(sin^2
2x/4)

split the fraction

since
sin^2 2x/2=2*sin^2 2x/4

The two fractions
become

(1/(sin^2 2x/4))
-2

times four to the first fraction top and
bottom

(4/sin^2 2x) -2

only
when sin^2 2x=1 does this function equal to two

This
question has some problem

The closest proof
is:

tan^2 x +cot^2 x=2 when x= 1/4pi + k/2pi
where k is a constant, a whole number