Is tan(x/2)=(tan(x))/((sec(x)+1)) ?

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

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prove tan(x/2)=(tan(x))/((sec(x)+1))

Solution

prove

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

Solution

True

Solution steps

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

Let:

u = \frac{x}{2}

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

Prove

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

True

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

Manipulating right side

\frac{tan ( 2 u )}{sec ( 2 u ) + 1}

Express with sin, cos

\frac{tan ( 2 u )}{1 + sec ( 2 u )}

Use the basic trigonometric identity:

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

= \frac{\frac{s i n ( 2 u )}{c o s ( 2 u )}}{1 + sec ( 2 u )}

Use the basic trigonometric identity:

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

= \frac{\frac{s i n ( 2 u )}{c o s ( 2 u )}}{1 + \frac{1}{c o s ( 2 u )}}

Simplify

\frac{\frac{s i n ( 2 u )}{c o s ( 2 u )}}{1 + \frac{1}{c o s ( 2 u )}} : \frac{sin ( 2 u )}{cos ( 2 u ) + 1}

\frac{\frac{s i n ( 2 u )}{c o s ( 2 u )}}{1 + \frac{1}{c o s ( 2 u )}}

Apply the fraction rule:

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

= \frac{sin ( 2 u )}{cos ( 2 u ) ( 1 + \frac{1}{c o s ( 2 u )} )}

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos (2 u) = cos (2 u)

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

= \frac{sin ( 2 u )}{\frac{c o s ( 2 u ) + 1}{c o s ( 2 u )} cos ( 2 u )}

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (2 u)

= cos (2 u) + 1

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

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

Use the Double Angle identity:

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

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

Simplify

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

\frac{2 sin ( u ) cos ( u )}{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{2 sin ( u ) cos ( u )}{2 cos ^{2} ( u )}

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

cos (u)

= \frac{sin ( u )}{cos ( u )}

= \frac{sin ( u )}{cos ( u )}

= \frac{sin ( u )}{cos ( u )}

Use the basic trigonometric identity:

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

= tan (u)

We showed that the two sides could take the same form

\Rightarrow True

Therefore

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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