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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Multiply

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

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

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

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

Multiply the numbers:

1 \cdot 2 = 2

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

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

Apply the fraction rule:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

sec (x) + tan (x) = \frac{1 + sin ( x )}{cos ( x )}

2 sin^{2} (2 x) + cos (4 x) = 1

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

cos (θ) - cos (θ) sin^{2} (θ) = cos^{3} (θ)

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

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