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

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

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Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating right side

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

Simplify

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

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

Least Common Multiplier of

1 + tan (x), 1 - tan (x) : (tan (x) + 1) (- tan (x) + 1)

1 + tan (x), 1 - tan (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

1 + tan (x)

1 - tan (x)

= (tan (x) + 1) (- tan (x) + 1)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

(tan (x) + 1) (- tan (x) + 1)

For

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

multiply the denominator and numerator by

- tan (x) + 1

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

For

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

multiply the denominator and numerator by

tan (x) + 1

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

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

Since the denominators are equal, combine the fractions:

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

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

Expand

- (- tan (x) + 1) + tan (x) + 1 : 2 tan (x)

- (- tan (x) + 1) + tan (x) + 1

- (- tan (x) + 1) : tan (x) - 1

- (- tan (x) + 1)

Distribute parentheses

= - (- tan (x)) - (1)

Apply minus-plus rules

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

= tan (x) - 1

= tan (x) - 1 + tan (x) + 1

Simplify

tan (x) - 1 + tan (x) + 1 : 2 tan (x)

tan (x) - 1 + tan (x) + 1

Group like terms

= tan (x) + tan (x) - 1 + 1

Add similar elements:

tan (x) + tan (x) = 2 tan (x)

= 2 tan (x) - 1 + 1

- 1 + 1 = 0

= 2 tan (x)

= 2 tan (x)

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

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

Rewrite using trig identities

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

Expand

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

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

Expand

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

(1 + tan (x)) (1 - tan (x))

Apply Difference of Two Squares Formula:

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

a = 1, b = tan (x)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - tan^{2} (x)

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

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

Use the Double Angle identity:

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

= tan (2 x)

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