trig <_<...........?

Question

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

Answer 1

Noting that tan² x + 1 = sec² x, RHS is given by :-

(tan² x + 1) / ( (2 - (tan² x + 1) )

= (tan² x + 1) / (1 - tan ² x)

= ((sin² x / cos² x) + 1)) / ((1 - (sin² x / cos² x))

=[(sin² x + cos² x) / cos² x] / [(cos² x - sin² x) / cos² x]

= (1 / cos² x ) / cos 2x / cos² x

= 1/cos 2x = sec 2x as required

Answer 2

We will start on the right hand side.

I will relabel for easy viewing:
Let S = sec(x) and C = cos(x).

So the right hand side becomes:
S^2 / (2 - S^2)

First recall that S = 1/C:
= (1/C)^2 / (2 - (1/C)^2)

Make the denominator into one big fraction:
= (1/C^2) / ((2C^2 - 1)/C^2)

Dividing two fractions is the same thing as multiplying the numerator by the reciprocal of the denominator:
= (1/C^2) * (C^2/(2C^2 - 1))

Notice that the C^2 cancels and we get:
= 1 / (2C^2 - 1)

Recall the identity: cos(2x) = 2(cos(x))^2 - 1
So in this case: cos(2x) = 2C^2 - 1:
= 1 / cos(2x)

Finally, we revert back to:
= sec(2x)

And this is the left hand side.

Answer 3

sec(2x)=1/cos(2x) : definition of secant.
=1/[2(cosx)^2-1] : Double-Angle Formula
=1/[2/(secx)^2-1] = 1/[2/(secx)^2-1] * ((secx)^2)/((secx)^2)) : definition of secant and multiplying 1 (secx/secx=1).

=(secx)^2/[2-(secx)^2] . (done)