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

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

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

prove

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

True

Solution steps

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

Manipulating right side

cot (x) cos (x)

Express with sin, cos

cos (x) cot (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

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

cos (x) cos (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

Rewrite using trig identities

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

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

We showed that the two sides could take the same form

\Rightarrow True

tan^{2} (x) = tan (2 x)

7 cos^{2} (x) + 5 sin^{2} (x) = cos (2 x) + 6

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

9 (1 + sin (y)) (1 + sin (- y)) = 9 cos^{2} (y)

(\frac{tan ( θ )}{sec ( θ )}) = sin (θ)

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