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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

(1 + cos (θ)) \cdot tan (\frac{θ}{2}) = sin (θ)

csc (θ) - cot (θ) = 1

5 (sin^{2} (x) - 1) = 5 cot^{2} (x)

\frac{sec ^{2} ( x )}{tan ( x )} = cot (x) + tan (x)

cos (z) = sin (\frac{π}{2} + z)

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