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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

= \frac{1}{1 + tan ( x ) \frac{2 t a n ( x )}{1 - t a n ^{2} ( x )}}

Simplify

\frac{1}{1 + tan ( x ) \frac{2 t a n ( x )}{1 - t a n ^{2} ( x )}} : \frac{1 - tan ^{2} ( x )}{1 + tan ^{2} ( x )}

\frac{1}{1 + tan ( x ) \frac{2 t a n ( x )}{1 - t a n ^{2} ( x )}}

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

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

Multiply fractions:

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

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

2 tan (x) tan (x) = 2 tan^{2} (x)

2 tan (x) tan (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

tan (x) tan (x) = tan^{1 + 1} (x)

= 2 tan^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= 2 tan^{2} (x)

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

= \frac{1}{1 + \frac{2 t a n ^{2} ( x )}{- t a n ^{2} ( x ) + 1}}

Join

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

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

Convert element to fraction:

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

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

1 \cdot (1 - tan^{2} (x)) + 2 tan^{2} (x) = 1 + tan^{2} (x)

1 \cdot (1 - tan^{2} (x)) + 2 tan^{2} (x)

1 \cdot (1 - tan^{2} (x)) = 1 - tan^{2} (x)

1 \cdot (1 - tan^{2} (x))

Multiply:

1 \cdot (1 - tan^{2} (x)) = (1 - tan^{2} (x))

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

Remove parentheses:

(a) = a

= 1 - tan^{2} (x)

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

Add similar elements:

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

= 1 + tan^{2} (x)

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

= \frac{1}{\frac{1 + t a n ^{2} ( x )}{1 - t a n ^{2} ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

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

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

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

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

Express with sin, cos

\frac{1 - tan ^{2} ( x )}{sec ^{2} ( x )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply exponent rule:

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

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

Apply the fraction rule:

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Cancel the common factor:

cos^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

= cos (2 x)

= cos (2 x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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