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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

2 tan (x) sec (x)

Express with sin, cos

2 sec (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply the numbers:

1 \cdot 2 = 2

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

Manipulating right side

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

Simplify

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

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

Least Common Multiplier of

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

1 + sin (x), 1 - sin (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

1 + sin (x)

1 - sin (x)

= (sin (x) + 1) (- sin (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

(sin (x) + 1) (- sin (x) + 1)

For

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

multiply the denominator and numerator by

- sin (x) + 1

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

For

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

multiply the denominator and numerator by

sin (x) + 1

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

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

Since the denominators are equal, combine the fractions:

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

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

Expand

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

- (- sin (x) + 1) + sin (x) + 1

- (- sin (x) + 1) : sin (x) - 1

- (- sin (x) + 1)

Distribute parentheses

= - (- sin (x)) - (1)

Apply minus-plus rules

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

= sin (x) - 1

= sin (x) - 1 + sin (x) + 1

Simplify

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

sin (x) - 1 + sin (x) + 1

Group like terms

= sin (x) + sin (x) - 1 + 1

Add similar elements:

sin (x) + sin (x) = 2 sin (x)

= 2 sin (x) - 1 + 1

- 1 + 1 = 0

= 2 sin (x)

= 2 sin (x)

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

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

Rewrite using trig identities

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

Expand

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

(1 + sin (x)) (1 - sin (x))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (x)

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

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (x)

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

Use the Pythagorean identity:

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

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

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

= \frac{2 sin ( 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 2tan(x)sec(x)= 1/(1-sin(x))-1/(1+sin(x)) ?
The answer to whether 2tan(x)sec(x)= 1/(1-sin(x))-1/(1+sin(x)) is True