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

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You want to solve this rather than prove it. It is not an identity.

`sin^2(x)+cos^2(x)=1` Pythagorean theorem

`tan^2(x)+1=sec^2(x)` and

`cot^2(x)+1=csc^2(x)`

`(sec^2(x)-1)+(csc^2(x)-1)=2`

`sec^2(x)+csc^2(x) - 2 = 2`

`sec^2(x)+csc^2(x) = 4`

`1/(cos^2(x))+1/(sin^2(x)) = 4`

`(cos^2(x)+sin^2(x))/(cos^2(x)sin^2(x)) = 4`

`1/(cos^2(x)sin^2(x))=4`

`1=4cos^2(x)sin^2(x)`

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

`sin^2(x)-sin^4(x)-1/4=0`

`sin^4(x)-sin^2(x)+1/4=0`

`(sin^2(x)-1/2)^2=0`

`sin^2(x)-1/2=0`

`sin^2(x)=1/2`

`sin(x)=+-sqrt(2)/2`

`x = pi/4 + npi/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**