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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

1 \cdot (1 - sin^{2} (x))

Multiply:

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

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

Remove parentheses:

(a) = a

= 1 - sin^{2} (x)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

cos (x) cot (x)

cos (x) cot (x)

We showed that the two sides could take the same form

\Rightarrow True

sin (x) + sin (x) cot^{2} (x) = csc (x)

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

cot^{2} (x) - cos^{2} (x) = cos^{2} (x) cot^{2} (x)

sin (\frac{π}{2} + x) = cos (x)

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

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