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

The answer to whether (1+csc(x))(1-sin(x))=cot(x)cos(x) is True

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prove (1+csc(x))(1-sin(x))=cot(x)cos(x)

prove

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

True

Solution steps

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

Manipulating left side

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

Rewrite using trig identities

\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 )}

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

cos (x) cot (x)

cos (x) cot (x)

= cot (x) cos (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( 5 x ) - sin ( x )}{cos ( 5 x ) - cos ( x )} = - cot (3 x)

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

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

tan (x) sec (x) (csc (x) - sin (x)) = 1

csc (x) cos (x) cot (x) = sin (x)

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