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

The answer to whether (1+csc(θ))(1-sin(θ))=csc(θ)-sin(θ) is True

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Popular >

prove (1+csc(θ))(1-sin(θ))=csc(θ)-sin(θ)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

(1 + csc (θ)) (1 - sin (θ))

Express with sin, cos

(1 + csc (θ)) (1 - sin (θ))

Use the basic trigonometric identity:

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

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

Simplify

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

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

Join

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

1 + \frac{1}{sin ( θ )}

Convert element to fraction:

1 = \frac{1 sin ( θ )}{sin ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot sin (θ) = sin (θ)

= \frac{sin ( θ ) + 1}{sin ( θ )}

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

Multiply fractions:

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

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

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

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

Expand

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

(1 + sin (θ)) (1 - sin (θ))

Apply Difference of Two Squares Formula:

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

a = 1, b = sin (θ)

= 1^{2} - sin^{2} (θ)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - sin^{2} (θ)

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

Manipulating right side

csc (θ) - sin (θ)

Express with sin, cos

csc (θ) - sin (θ)

Use the basic trigonometric identity:

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

= \frac{1}{sin ( θ )} - sin (θ)

Simplify

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

\frac{1}{sin ( θ )} - sin (θ)

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

1 - sin (θ) sin (θ) = 1 - sin^{2} (θ)

1 - sin (θ) sin (θ)

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

= 1 - sin^{2} (θ)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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