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

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

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

prove

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

True

Solution steps

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

Manipulating right side

\frac{tan ( x )}{1 + tan ^{2} ( x )}

Express with sin, cos

\frac{tan ( x )}{1 + tan ^{2} ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

= \frac{sin ( x )}{cos ( x ) ( 1 + ( \frac{s i n ( x )}{c o s ( x )} ) ^{2} )}

Apply exponent rule:

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

= \frac{sin ( x )}{cos ( x ) ( \frac{s i n ^{2} ( x )}{c o s ^{2} ( x )} + 1 )}

Join

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

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

Convert element to fraction:

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

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

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} ( x ) + sin ^{2} ( x )}{cos ^{2} ( x )}

Multiply:

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

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

= \frac{sin ( x )}{\frac{c o s ^{2} ( x ) + s i n ^{2} ( x )}{c o s ^{2} ( x )} cos ( x )}

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

= \frac{sin ( x )}{\frac{c o s ^{2} ( x ) + s i n ^{2} ( x )}{c o s ( x )}}

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

= \frac{\frac{s i n ( 2 x )}{2}}{cos ^{2} ( x ) + sin ^{2} ( x )}

Apply the fraction rule:

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

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

Use the Pythagorean identity:

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

= \frac{sin ( 2 x )}{2 \cdot 1}

Simplify

= \frac{sin ( 2 x )}{2}

= \frac{sin ( 2 x )}{2}

We showed that the two sides could take the same form

\Rightarrow True

p ro v e csc (\frac{π}{12}) = sec (\frac{5 π}{12})

p ro v e 1 - \frac{cot ^{2} ( θ )}{csc ^{2} ( θ )} = sin^{2} (θ)

p ro v e csc^{2} (t) tan^{2} (t) - 1 = tan^{2} (t)

p ro v e \frac{1}{2} cos (x) = \frac{cos ( x )}{2}

p ro v e \frac{sec ^{2} ( θ ) - 1}{sin ^{2} ( θ )} = sec^{2} (θ)

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