2x=sec theta and 2/x=tan theta find the value of 2(x^2 - 1/x^2)

Answer 1

Using the definitions of tan and sec in terms of
opposite, adjacent and hypotenuse ...

tanØ = 2/x = opp/adj
so we have a right-angled triangle with base angle Ø, opposite as 2 and adjacent as x
by Pythagoras,
r , (the hypotenuse) = (x^2 + 4)^(1/2 or √(x^2 + 4)
and secØ = √(x^2 + 4)/x

but we are given that secØ = 2x , so ....

2x = √(x^2 + 4)/x
2x^2 = √(x^2 + 4)
square both sides:
4x^4 = x^2 + 4
4x^4 - x^2 - 4 = 0
x^2 = (1 ± √(1+64)/8
x^2 = (1 + √65)/8 , rejecting the negative value since x^2 can't be negative

so 2(x^2 - 1/x^2)
= 2( (1+√65)/8) - (8/(1+√65) )
= 0.5

Answer 2

sec^2 = 1+tan^2 so
(2x)^2 = 1+(2/x)^2
4x^2 = 1 + 4/x^2
4x^4 - x^2 - 4 = 0
and proceed as above.

Or, forget about the value of x altogether.

secØ = 2x, so x = secØ/2
tanØ = 2/x, so 1/x = tanØ/2

2(x^2-1/x^2)
2(sec^2Ø/4 - tan^2Ø/4)
1/2(sec^2Ø - tan^2Ø)
1/2