Is sec^2(x)= 2/(1+cos(2x)) ?

The answer to whether sec^2(x)= 2/(1+cos(2x)) is True

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prove sec^2(x)= 2/(1+cos(2x))

prove

sec^{2} (x) = \frac{2}{1 + cos ( 2 x )}

True

Solution steps

sec^{2} (x) = \frac{2}{1 + cos ( 2 x )}

Manipulating right side

\frac{2}{1 + cos ( 2 x )}

Rewrite using trig identities

\frac{2}{1 + cos ( 2 x )}

Use the Double Angle identity:

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

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

Simplify

\frac{2}{1 + 2 cos ^{2} ( x ) - 1} : \frac{1}{cos ^{2} ( x )}

\frac{2}{1 + 2 cos ^{2} ( x ) - 1}

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

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

Group like terms

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

1 - 1 = 0

= 2 cos^{2} (x)

= \frac{2}{2 cos ^{2} ( x )}

Divide the numbers:

\frac{2}{2} = 1

= \frac{1}{cos ^{2} ( x )}

= \frac{1}{cos ^{2} ( x )}

= \frac{1}{cos ^{2} ( x )}

Rewrite using trig identities

Use the basic trigonometric identity:

cos (x) = \frac{1}{sec ( x )}

\frac{1}{( \frac{1}{s e c ( x )} ) ^{2}}

Simplify

\frac{1}{( \frac{1}{s e c ( x )} ) ^{2}}

(\frac{1}{sec ( x )})^{2} = \frac{1}{sec ^{2} ( x )}

(\frac{1}{sec ( x )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= \frac{1 ^{2}}{sec ^{2} ( x )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( x )}

= \frac{1}{\frac{1}{s e c ^{2} ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{sec ^{2} ( x )}{1}

Apply rule

\frac{a}{1} = a

= sec^{2} (x)

sec^{2} (x)

sec^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

sin (2 θ) = 2 sin^{2} (θ) - 1

sec (x) - sin (x) = cot (x) cos (x)

\frac{csc ( A )}{csc ( A ) - sin ( A )} = sec^{2} (A)

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

sec (x) = sin (x) tan (x) = cos (x)

Is sec^2(x)= 2/(1+cos(2x)) ?
The answer to whether sec^2(x)= 2/(1+cos(2x)) is True