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

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

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Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Simplify

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

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

Least Common Multiplier of

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

1 + tan (θ), 1 - tan (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

1 + tan (θ)

1 - tan (θ)

= (tan (θ) + 1) (- tan (θ) + 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 (θ) + 1) (- tan (θ) + 1)

For

\frac{1}{1 + tan ( θ )} :

multiply the denominator and numerator by

- tan (θ) + 1

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

For

\frac{1}{1 - tan ( θ )} :

multiply the denominator and numerator by

tan (θ) + 1

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

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

Since the denominators are equal, combine the fractions:

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

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

Expand

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

- (- tan (θ) + 1) + tan (θ) + 1

- (- tan (θ) + 1) : tan (θ) - 1

- (- tan (θ) + 1)

Distribute parentheses

= - (- tan (θ)) - (1)

Apply minus-plus rules

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

= tan (θ) - 1

= tan (θ) - 1 + tan (θ) + 1

Simplify

tan (θ) - 1 + tan (θ) + 1 : 2 tan (θ)

tan (θ) - 1 + tan (θ) + 1

Group like terms

= tan (θ) + tan (θ) - 1 + 1

Add similar elements:

tan (θ) + tan (θ) = 2 tan (θ)

= 2 tan (θ) - 1 + 1

- 1 + 1 = 0

= 2 tan (θ)

= 2 tan (θ)

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

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

Rewrite using trig identities

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

Expand

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

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

Expand

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

(1 + tan (θ)) (1 - tan (θ))

Apply Difference of Two Squares Formula:

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

a = 1, b = tan (θ)

= 1^{2} - tan^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - tan^{2} (θ)

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

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

Use the Double Angle identity:

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

= tan (2 θ)

= tan (2 θ)

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