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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Multiply:

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

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

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

Apply rule

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

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

cos (3 x) + cos (x) = 2 cos (2 x) cos (x)

sin (4 x) = 2 sin (2 x) cos (2 x)

tan (x) csc^{2} (x) - tan (x) = cot (x)

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

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