Is tan(x/2)=(sin(x))/(cos(x)+1) ?

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

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prove tan(x/2)=(sin(x))/(cos(x)+1)

prove

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

True

Solution steps

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

Let:

u = \frac{x}{2}

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

Prove

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

True

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

Manipulating right side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

- 1 + 1 = 0

= \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{sin ( x )}{cos ( x ) + 1}

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( x ) csc ( x )}{cos ( x )} = sec (x)

tan (x) + sec (x) = 1

- sinh^{2} (x) + cosh^{2} (x) = 1

sin^{2} (x) + cos^{2} (x) - 1 = 0

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

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