Is csc(θ)-csc(θ)cos^2(θ)=sin(θ) ?

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

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prove csc(θ)-csc(θ)cos^2(θ)=sin(θ)

prove

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

True

Solution steps

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

Manipulating left side

csc (θ) - csc (θ) cos^{2} (θ)

Express with sin, cos

csc (θ) - cos^{2} (θ) csc (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

cos^{2} (θ) \frac{1}{sin ( θ )}

Multiply fractions:

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

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

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

= \frac{cos ^{2} ( θ )}{sin ( θ )}

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

Apply rule

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

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

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

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

Rewrite using trig identities

\frac{1 - cos ^{2} ( θ )}{sin ( θ )}

Use the Pythagorean identity:

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

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

= \frac{sin ^{2} ( θ )}{sin ( θ )}

Cancel the common factor:

sin (θ)

= sin (θ)

= sin (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( 6 x ) + sin ( 2 x )}{cos ( 6 x ) + cos ( 2 x )} = tan (4 x)

sec^{2} (a) = 1 + tan^{2} (a)

\frac{tan ( θ )}{csc ( θ )} = sec (θ) - cos (θ)

cot (2 β) = \frac{cot ^{2} ( β ) - 1}{2 cot ( β )}

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

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