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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Multiply by

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

= \frac{cos ( 2 x )}{\frac{( 1 - s i n ( 2 x ) ) ( 1 + s i n ( 2 x ) )}{1 + s i n ( 2 x )}}

Apply Difference of Two Squares Formula:

2 x^{2} - y^{2} = (2 x + y) (2 x - y)

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

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Apply the fraction rule:

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

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

Cancel the common factor:

cos (2 x)

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

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

Use the Double Angle identity:

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

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

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

Add similar elements:

- sin^{2} (x) + 2 sin^{2} (x) = sin^{2} (x)

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

Apply Perfect Square Formula:

(a + b)^{2} = a^{2} + 2 ab + b^{2}

a = sin (x), b = cos (x)

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

Cancel the common factor:

sin (x) + cos (x)

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

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

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

Manipulating right side

\frac{1 + tan ( x )}{1 - tan ( x )}

Express with sin, cos

\frac{1 + tan ( x )}{1 - tan ( x )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

1 - \frac{sin ( x )}{cos ( x )}

Convert element to fraction:

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

= \frac{1 \cdot cos ( x )}{cos ( x )} - \frac{sin ( x )}{cos ( 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 ( x ) - sin ( x )}{cos ( x )}

Multiply:

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

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

Divide fractions:

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

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

Cancel the common factor:

cos (x)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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