Can someone please help me with a trig proof? sec2x = (sec^2x)/(2-sec^2x)?

Question

I've tried forever, and I feel stupid, but I can't get the answer. Any help would be greatly appreciated.

Answer 1

sec(2x) = [sec^2(x)] / [2 - sec^2(x)]

Your first step would be to choose the more complex side. I'm going to choose the right hand side (RHS).

RHS = [sec^2(x)] / [2 - sec^2(x)]

Now, I'm going to convert everything to sines and cosines.

RHS = [1/cos^2(x)] / [2 - 1/cos^2(x)]

Multiplying the top and bottom by cos^2(x), we get

RHS = 1 / [2cos^2(x) - 1]

There's an identity out there called the half angle identity, which goes as follows:

cos(2y) = cos^2(y) - sin^2(y) = 1 - 2sin^2(y) = 2cos^2(y) - 1

Using the third form of the identity,

RHS = 1 / cos(2x)

Which, by definition is equal to

RHS = 1 / cos(2x) = sec(2x) = LHS

Answer 2

Start by rewritting the secant function in terms of the cosine function.
sec(2x)=1/cos(2x)
Then use the trigonometric identity cos(2x) = cos^2(x) - sin^2(x) = 2 cos^2(x) - 1 = 1 - 2 sin^2(x).

For the right side, sec^2x =1/cos^2x. Then use the trig identity of cos^2(x) = 1/2 + 1/2 cos(2x).

Essentially just use the above and whole lot of algebra to get the same expression on each side.

Answer 3

Prove the identity.

sec2x = (sec² x)/(2 - sec² x)

Let's work with the right hand side (RHS) to show it is the same as the left.

Remember sec x = 1/cos x.
Remember also the double angle formula cos(2x) = 2cos² x - 1.

RHS = (sec² x)/(2 - sec² x)
Now multiply numerator and denominator by cos² x.

= 1/(2cos²x - 1)
Now apply the double angle formula.

= 1/cos(2x)
= sec(2x) = left hand side.

Answer 4

Start with the right-hand side:
(sec^2x)/(2 - sec^2x) = 1/(cos^2x) * 1/(2 - 1/(cos^2x))
Multiply the two fractions:
= 1 / (2cos^2x - 1)
Note that 1 = cos^2x + sin^2x:
= 1 / (cos^2x + cos^2x - cos^2x - sin^2x)
= 1 / (cos^2x - sin^2x)
= 1 / (cos2x)
= sec2x

QED

Answer 5

Soul Train

Answer 6

I am super math stupid but you can find help with this by plugging the equation into a google window and hit return.

Answer 7

sec(2x) = (sec(x)^2)/(2 - sec(x)^2)

sec(2x) = 1/(cos(2x)) =
1/(cos(x)^2 - sin(x)^2), 1/(2cos(x)^2 - 1), or 1/(1 - 2sin(x)^2)

(sec(x)^2)/(2 - sec(x)^2)
(1/cos(x)^2)/(2 - (1/cos(x)^2))
(1/cos(x)^2)/((2cos(x)^2 - 1)/(cos(x)^2))
(1/cos(x)^2) * ((cos(x)^2)/(2cos(x)^2 - 1))
(cos(x)^2)/((cos(x)^2)(2cos(x)^2 - 1))
1/(2cos(x)^2 - 1)
sec(2x)

so

sec(2x) = (sec(x)^2)/(2 - sec(x)^2)