Is (1-sin^2(x))csc^2(x)=cot^2(x) ?

The answer to whether (1-sin^2(x))csc^2(x)=cot^2(x) is True

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prove (1-sin^2(x))csc^2(x)=cot^2(x)

prove

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

True

Solution steps

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

Manipulating right side

cot^{2} (x)

Express with sin, cos

cot^{2} (x)

Use the basic trigonometric identity:

cot (x) = \frac{cos ( x )}{sin ( x )}

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

Simplify

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

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

Apply exponent rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Use the basic trigonometric identity:

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

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

We showed that the two sides could take the same form

\Rightarrow True

p ro v e sin (t) tan (t) = \frac{1 - cos ^{2} ( t )}{cos ( t )}

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

p ro v e \frac{csc ( x ) - cot ( x )}{sec ( x ) - 1} = cot (x)

p ro v e \frac{1 + cos ( x )}{1 - cos ( x )} - \frac{1 - cos ( x )}{1 + cos ( x )} = 4 cot (x) csc (x)

p ro v e cos (z + w) = cos (z) cos (w) - sin (z) sin (w)

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