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

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

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Popular Trigonometry >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Multiply

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

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

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

cos (x)

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

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

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

Use the basic trigonometric identity:

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

= cot (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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