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

csc^{2} (θ) (1 - cos^{2} (θ)) = sin (648 0^{\circ})

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

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

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

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

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