Is 2-2tan(x)cot(2x)=sec^2(x) ?

The answer to whether 2-2tan(x)cot(2x)=sec^2(x) is True

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prove 2-2tan(x)cot(2x)=sec^2(x)

Solution

prove

2 - 2 tan (x) cot (2 x) = sec^{2} (x)

Solution

True

Solution steps

2 - 2 tan (x) cot (2 x) = sec^{2} (x)

Manipulating left side

2 - 2 tan (x) cot (2 x)

Express with sin, cos

2 - 2 cot (2 x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply

2 \cdot \frac{cos ( 2 x )}{sin ( 2 x )} \cdot \frac{sin ( x )}{cos ( x )} : \frac{2 cos ( 2 x ) sin ( x )}{sin ( 2 x ) cos ( x )}

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

Multiply fractions:

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

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

Use the Double Angle identity:

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

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

Simplify

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

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

- 2 (2 cos^{2} (x) - 1) sin (x) + 2 cos (x) \cdot 2 sin (x) cos (x) = - 2 sin (x) (2 cos^{2} (x) - 1) + 4 cos^{2} (x) sin (x)

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

2 cos (x) \cdot 2 sin (x) cos (x) = 4 cos^{2} (x) sin (x)

2 cos (x) \cdot 2 sin (x) cos (x)

Multiply the numbers:

2 \cdot 2 = 4

= 4 cos (x) sin (x) cos (x)

Apply exponent rule:

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

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

= 4 sin (x) cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= 4 sin (x) cos^{2} (x)

= - 2 sin (x) (2 cos^{2} (x) - 1) + 4 cos^{2} (x) sin (x)

= \frac{- 2 sin ( x ) ( 2 cos ^{2} ( x ) - 1 ) + 4 cos ^{2} ( x ) sin ( x )}{2 cos ( x ) sin ( x ) cos ( x )}

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

cos (x) \cdot 2 sin (x) cos (x)

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

= \frac{- 2 sin ( x ) ( 2 cos ^{2} ( x ) - 1 ) + 4 cos ^{2} ( x ) sin ( x )}{2 cos ^{2} ( x ) sin ( x )}

Factor

- 2 (2 cos^{2} (x) - 1) sin (x) + 4 sin (x) cos^{2} (x) : 2 sin (x)

- 2 (2 cos^{2} (x) - 1) sin (x) + 4 sin (x) cos^{2} (x)

Rewrite as

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

Factor out common term

2 sin (x)

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

Expand

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

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

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

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

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

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

Add similar elements:

- 2 cos^{2} (x) + 2 cos^{2} (x) = 0

= 1

= 2 \cdot 1 \cdot sin (x)

Refine

= 2 sin (x)

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

Cancel

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

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

Divide the numbers:

\frac{2}{2} = 1

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

Cancel the common factor:

sin (x)

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

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{( \frac{1}{s e c ( x )} ) ^{2}}

Simplify

\frac{1}{( \frac{1}{s e c ( x )} ) ^{2}}

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{1}{\frac{1}{s e c ^{2} ( x )}}

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec^{2} (x)

sec^{2} (x)

sec^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is 2-2tan(x)cot(2x)=sec^2(x) ?
The answer to whether 2-2tan(x)cot(2x)=sec^2(x) is True