Is sec(2pi)=sqrt(1-tan^2(2pi)) ?

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

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prove sec(2pi)=sqrt(1-tan^2(2pi))

prove

sec (2 π) = 1 - tan^{2} (2 π)

True

Solution steps

sec (2 π) = 1 - tan^{2} (2 π)

Manipulating left side

sec (2 π)

Simplify

sec (2 π) : 1

sec (2 π)

sec (2 π) = sec (0)

sec (2 π)

Rewrite

2 π

2 π + 0

= sec (2 π + 0)

Apply the periodicity of

sec

sec (x + 2 π) = sec (x)

sec (2 π + 0) = sec (0)

= sec (0)

= sec (0)

Use the following trivial identity:

sec (0) = 1

sec (0)

sec (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

= 1

= 1

= 1

Manipulating right side

1 - tan^{2} (2 π)

Simplify

1 - tan^{2} (2 π) : 1

1 - tan^{2} (2 π)

tan^{2} (2 π) = 0

tan^{2} (2 π)

tan (2 π) = 0

tan (2 π)

tan (2 π) = tan (0)

tan (2 π)

Rewrite

2 π

π \cdot 2 + 0

= tan (π 2 + 0)

Apply the periodicity of

tan

tan (x + π \cdot k) = tan (x)

tan (π \cdot 2 + 0) = tan (0)

= tan (0)

= tan (0)

Use the following trivial identity:

tan (0) = 0

tan (0)

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 0

= 0

= 0^{2}

Apply rule

0^{a} = 0

= 0

= 1 - 0

Simplify

1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

tan^{2} (θ) cos^{2} (θ) = 1 - cos^{2} (θ)

\frac{sin ( x )}{sin ( 2 x )} = \frac{sec ( x )}{2}

sin (2 x) = 2 sin (x) \cdot cos (x)

tan (a) \frac{1}{sin ^{2} ( a )} = csc (a) sec (a)

(1 + csc (x)) (1 + csc (- x)) = - cot^{4} (x)

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