how to prove tan^2x+cot^2x=2 sec^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`
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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
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