Is tan(A+B)+tan(A-B)=((2sin(2A)))/((cos(2A)+cos(2B))) ?

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

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Popular >

prove tan(A+B)+tan(A-B)=((2sin(2A)))/((cos(2A)+cos(2B)))

Solution

prove

tan (A + B) + tan (A - B) = \frac{( 2 sin ( 2 A ) )}{( cos ( 2 A ) + cos ( 2 B ) )}

Solution

True

Solution steps

tan (A + B) + tan (A - B) = \frac{2 sin ( 2 A )}{cos ( 2 A ) + cos ( 2 B )}

Manipulating left side

tan (A + B) + tan (A - B)

Express with sin, cos

tan (A + B) + tan (A - B)

Use the basic trigonometric identity:

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

= \frac{sin ( A + B )}{cos ( A + B )} + tan (A - B)

Use the basic trigonometric identity:

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

= \frac{sin ( A + B )}{cos ( A + B )} + \frac{sin ( A - B )}{cos ( A - B )}

Simplify

\frac{sin ( A + B )}{cos ( A + B )} + \frac{sin ( A - B )}{cos ( A - B )} : \frac{sin ( A + B ) cos ( A - B ) + sin ( A - B ) cos ( A + B )}{cos ( A + B ) cos ( A - B )}

\frac{sin ( A + B )}{cos ( A + B )} + \frac{sin ( A - B )}{cos ( A - B )}

Least Common Multiplier of

cos (A + B), cos (A - B) : cos (A + B) cos (A - B)

cos (A + B), cos (A - B)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

cos (A + B)

cos (A - B)

= cos (A + B) cos (A - B)

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

cos (A + B) cos (A - B)

For

\frac{sin ( A + B )}{cos ( A + B )} :

multiply the denominator and numerator by

cos (A - B)

\frac{sin ( A + B )}{cos ( A + B )} = \frac{sin ( A + B ) cos ( A - B )}{cos ( A + B ) cos ( A - B )}

For

\frac{sin ( A - B )}{cos ( A - B )} :

multiply the denominator and numerator by

cos (A + B)

\frac{sin ( A - B )}{cos ( A - B )} = \frac{sin ( A - B ) cos ( A + B )}{cos ( A - B ) cos ( A + B )}

= \frac{sin ( A + B ) cos ( A - B )}{cos ( A + B ) cos ( A - B )} + \frac{sin ( A - B ) cos ( A + B )}{cos ( A - B ) cos ( A + B )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( A + B ) cos ( A - B ) + sin ( A - B ) cos ( A + B )}{cos ( A + B ) cos ( A - B )}

= \frac{sin ( A + B ) cos ( A - B ) + sin ( A - B ) cos ( A + B )}{cos ( A + B ) cos ( A - B )}

= \frac{cos ( A + B ) sin ( A - B ) + cos ( A - B ) sin ( A + B )}{cos ( A + B ) cos ( A - B )}

Rewrite using trig identities

\frac{cos ( A + B ) sin ( A - B ) + cos ( A - B ) sin ( A + B )}{cos ( A + B ) cos ( A - B )}

Use the Product to Sum identity:

cos (s) cos (t) = \frac{1}{2} (cos (s - t) + cos (s + t))

= \frac{cos ( A + B ) sin ( A - B ) + cos ( A - B ) sin ( A + B )}{\frac{1}{2} ( cos ( A + B - ( A - B ) ) + cos ( A + B + A - B ) )}

Simplify

\frac{cos ( A + B ) sin ( A - B ) + cos ( A - B ) sin ( A + B )}{\frac{1}{2} ( cos ( A + B - ( A - B ) ) + cos ( A + B + A - B ) )} : \frac{2 ( cos ( A + B ) sin ( A - B ) + cos ( A - B ) sin ( A + B ) )}{cos ( 2 B ) + cos ( 2 A )}

\frac{cos ( A + B ) sin ( A - B ) + cos ( A - B ) sin ( A + B )}{\frac{1}{2} ( cos ( A + B - ( A - B ) ) + cos ( A + B + A - B ) )}

Simplify

A + B + A - B : 2 A

A + B + A - B

Group like terms

= A + A + B - B

Add similar elements:

A + A = 2 A

= 2 A + B - B

Add similar elements:

B - B = 0

= 2 A

= \frac{cos ( A + B ) sin ( A - B ) + cos ( A - B ) sin ( A + B )}{\frac{1}{2} ( cos ( A + B - ( A - B ) ) + cos ( 2 A ) )}

Multiply

\frac{1}{2} (cos (A + B - (A - B)) + cos (2 A)) : \frac{cos ( 2 B ) + cos ( 2 A )}{2}

\frac{1}{2} (cos (A + B - (A - B)) + cos (2 A))

Multiply fractions:

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

= \frac{1 \cdot ( cos ( A + B - ( A - B ) ) + cos ( 2 A ) )}{2}

1 \cdot (cos (A + B - (A - B)) + cos (2 A)) = cos (A + B - (A - B)) + cos (2 A)

1 \cdot (cos (A + B - (A - B)) + cos (2 A))

Multiply:

1 \cdot (cos (A + B - (A - B)) + cos (2 A)) = (cos (A + B - (A - B)) + cos (2 A))

= (cos (A + B - (A - B)) + cos (2 A))

Remove parentheses:

(a) = a

= cos (A + B - (A - B)) + cos (2 A)

= \frac{cos ( A + B - ( A - B ) ) + cos ( 2 A )}{2}

Expand

A + B - (A - B) : 2 B

A + B - (A - B)

- (A - B) : - A + B

- (A - B)

Distribute parentheses

= - (A) - (- B)

Apply minus-plus rules

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

= - A + B

= A + B - A + B

Simplify

A + B - A + B : 2 B

A + B - A + B

Group like terms

= A - A + B + B

Add similar elements:

A - A = 0

= B + B

Add similar elements:

B + B = 2 B

= 2 B

= 2 B

= \frac{cos ( 2 B ) + cos ( 2 A )}{2}

= \frac{cos ( A + B ) sin ( A - B ) + cos ( A - B ) sin ( A + B )}{\frac{c o s ( 2 B ) + c o s ( 2 A )}{2}}

Apply the fraction rule:

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

= \frac{( cos ( A + B ) sin ( A - B ) + cos ( A - B ) sin ( A + B ) ) \cdot 2}{cos ( 2 B ) + cos ( 2 A )}

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

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

Rewrite as

= sin (A + B) cos (A - B) + cos (A + B) sin (A - B)

Rewrite using trig identities

sin (A + B) cos (A - B) + cos (A + B) sin (A - B)

Use the Angle Sum identity:

sin (s) cos (t) + cos (s) sin (t) = sin (s + t)

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

Simplify

A + B + A - B : 2 A

A + B + A - B

Group like terms

= A + A + B - B

Add similar elements:

A + A = 2 A

= 2 A + B - B

Add similar elements:

B - B = 0

= 2 A

= \frac{2 sin ( 2 A )}{cos ( 2 B ) + cos ( 2 A )}

= \frac{2 sin ( 2 A )}{cos ( 2 B ) + cos ( 2 A )}

= \frac{2 sin ( 2 A )}{cos ( 2 A ) + cos ( 2 B )}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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