Is 1=(sec(2x))^2-(tan(2x))^2 ?

The answer to whether 1=(sec(2x))^2-(tan(2x))^2 is True

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prove 1=(sec(2x))^2-(tan(2x))^2

prove

1 = (sec (2 x))^{2} - (tan (2 x))^{2}

True

Solution steps

1 = (sec (2 x))^{2} - (tan (2 x))^{2}

Manipulating right side

(sec (2 x))^{2} - (tan (2 x))^{2}

Simplify

= sec^{2} (2 x) - tan^{2} (2 x)

Rewrite using trig identities

sec^{2} (2 x) - tan^{2} (2 x)

Use the Pythagorean identity:

sec^{2} (x) = tan^{2} (x) + 1

sec^{2} (x) - tan^{2} (x) = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

2 tan (x) = \frac{2 sin ( x )}{cos ( x )}

(sin (x) + cos (x))^{2} - 1 = sin (2 x)

sin (t) cos (t) = 1

\frac{1}{1 - sin ( B )} = sec^{2} (B) + sec (B) tan (B)

tan (6 0^{\circ}) = \frac{2 tan ( 3 0 ^{\circ} )}{1 - tan ^{2} ( 3 0 ^{\circ} )}

Is 1=(sec(2x))^2-(tan(2x))^2 ?
The answer to whether 1=(sec(2x))^2-(tan(2x))^2 is True