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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Join

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

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

Multiply:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (2 x)

= cos (2 x) + 1

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

cos (x)

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

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

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

Use the basic trigonometric identity:

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

= tan (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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