Is (1-sin(x))(1+csc(x))=csc(x)-sin(x) ?

The answer to whether (1-sin(x))(1+csc(x))=csc(x)-sin(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-sin(x))(1+csc(x))=csc(x)-sin(x)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Simplify

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

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

Join

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (x) = sin (x)

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

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

Multiply fractions:

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

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

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

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

Manipulating right side

csc (x) - sin (x)

Express with sin, cos

csc (x) - sin (x)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

1 - sin (x) sin (x)

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

= 1 - sin^{2} (x)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

Popular Examples

prove 1-cos^2(θ)=sin^2(θ)prove

1 - cos^{2} (θ) = sin^{2} (θ)

prove tan(pi/4-θ)=(cos(θ)-sin(θ))/(cos(θ)+sin(θ))prove

tan (\frac{π}{4} - θ) = \frac{cos ( θ ) - sin ( θ )}{cos ( θ ) + sin ( θ )}

prove 2cos^2(x)+sin^2(x)=1+cos^2(x)prove

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

prove csc(B)-sin(B)=cot(B)cos(B)prove

csc (B) - sin (B) = cot (B) cos (B)

prove (1+sec^2(θ))/(sec^2(θ))=1+cos^2(θ)prove

\frac{1 + sec ^{2} ( θ )}{sec ^{2} ( θ )} = 1 + cos^{2} (θ)

Frequently Asked Questions (FAQ)

Is (1-sin(x))(1+csc(x))=csc(x)-sin(x) ?
The answer to whether (1-sin(x))(1+csc(x))=csc(x)-sin(x) is True