Is (sin(θ))/(1-cos^2(θ))=csc(θ) ?

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

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prove (sin(θ))/(1-cos^2(θ))=csc(θ)

prove

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

True

Solution steps

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

Manipulating left side

\frac{sin ( θ )}{1 - cos ^{2} ( θ )}

Rewrite using trig identities

\frac{sin ( θ )}{1 - cos ^{2} ( θ )}

Use the Pythagorean identity:

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

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

= \frac{sin ( θ )}{sin ^{2} ( θ )}

Cancel the common factor:

sin (θ)

= \frac{1}{sin ( θ )}

= \frac{1}{sin ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{c s c ( θ )}}

Simplify

\frac{1}{\frac{1}{c s c ( θ )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{csc ( θ )}{1}

Apply rule

\frac{a}{1} = a

= csc (θ)

csc (θ)

csc (θ)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e cos (x - \frac{π}{2}) = cos (x) tan (x)

p ro v e csc (- θ) = - csc (θ)

p ro v e sec (x) - tan (x) (sin (x)) = \frac{1}{sec ( x )}

p ro v e \frac{1 - tan ^{4} ( x )}{sec ^{2} ( x )} = 1 - tan^{2} (x)

p ro v e 3 - sec^{2} (x) = 2 - tan^{2} (x)

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