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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left 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 )}

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

\frac{2 tan ( x )}{cos ( x )}

\frac{2 tan ( x )}{cos ( x )}

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove cot(b)sin(b)=cos(b)prove

cot (b) sin (b) = cos (b)

prove cos(t)cot(t)=(1-sin^2(t))/(sin(t))prove

cos (t) cot (t) = \frac{1 - sin ^{2} ( t )}{sin ( t )}

prove csc(2x)=(1+cot^2(x))/(2cot(x))prove

csc (2 x) = \frac{1 + cot ^{2} ( x )}{2 cot ( x )}

prove (1-cos(θ))(1+cos(θ))(1+cot^2(θ))=1 prove

(1 - cos (θ)) (1 + cos (θ)) (1 + cot^{2} (θ)) = 1

prove tan^2(θ)-sin^2(θ)=sin^2(θ)tan^2(θ)prove

tan^{2} (θ) - sin^{2} (θ) = sin^{2} (θ) tan^{2} (θ)

Frequently Asked Questions (FAQ)

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