Is 1/(csc^2(x))=1-cos^2(x) ?

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

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prove 1/(csc^2(x))=1-cos^2(x)

prove

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

True

Solution steps

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

Manipulating left side

\frac{1}{csc ^{2} ( x )}

Express with sin, cos

\frac{1}{csc ^{2} ( x )}

Use the basic trigonometric identity:

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

= \frac{1}{( \frac{1}{s i n ( x )} ) ^{2}}

Simplify

\frac{1}{( \frac{1}{s i n ( x )} ) ^{2}} : sin^{2} (x)

\frac{1}{( \frac{1}{s i n ( x )} ) ^{2}}

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{1}{\frac{1}{s i n ^{2} ( x )}}

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sin^{2} (x)

= sin^{2} (x)

= sin^{2} (x)

Rewrite using trig identities

sin^{2} (x)

Use the Pythagorean identity:

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

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

= 1 - cos^{2} (x)

= 1 - cos^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

p ro v e tan^{2} (x) - (sec (x) + 1) (sec (x) - 1) = 0

p ro v e sec^{4} (θ) = sec^{2} (θ) tan^{2} (θ) + sec^{2} (θ)

p ro v e sin^{2} (θ) = 1 - \frac{1}{sec ^{2} ( θ )}

p ro v e tan (θ) sec (θ) csc (θ) = 1 + tan^{2} (θ)

p ro v e \frac{1 - cos ( x )}{\frac{1}{c o s ( x )} - 1} = cos (x)

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