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

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

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

prove

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

True

Solution steps

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

Let:

u = \frac{x}{2}

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

Prove

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

True

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

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

Group like terms

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

1 - 1 = 0

= 2 cos^{2} (u)

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply

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

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

Multiply fractions:

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

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

Multiply the numbers:

1 \cdot 2 = 2

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

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

Apply the fraction rule:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

Therefore

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

We showed that the two sides could take the same form

\Rightarrow True

cos (x) = \frac{15}{17}

tan (b) + cot (b) = \frac{1}{sin ( b ) cos ( b )}

csc (A) - sin (A) = (cos (A)) (cot (A))

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

\frac{( tan ( x ) cot ( x ) )}{sec ( x )} = cos (x)

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