Is (1-cos^2(x))(csc(x))=sin(x) ?

The answer to whether (1-cos^2(x))(csc(x))=sin(x) is True

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prove (1-cos^2(x))(csc(x))=sin(x)

prove

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

True

Solution steps

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

Manipulating left side

(1 - cos^{2} (x)) csc (x)

Express with sin, cos

(1 - cos^{2} (x)) csc (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

1 \cdot (1 - cos^{2} (x)) = 1 - cos^{2} (x)

1 \cdot (1 - cos^{2} (x))

Multiply:

1 \cdot (1 - cos^{2} (x)) = (1 - cos^{2} (x))

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

Remove parentheses:

(a) = a

= 1 - cos^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

sin (x)

= sin (x)

= sin (x)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( x + y )}{sin ( x ) cos ( y )} = 1 + cot (x) tan (y)

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

sin^{2} (2 x) = \frac{1}{2} - \frac{1}{2} cos (4 x)

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

sin (\frac{a}{2}) = \frac{1}{2} sin (a)

Is (1-cos^2(x))(csc(x))=sin(x) ?
The answer to whether (1-cos^2(x))(csc(x))=sin(x) is True