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

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

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

prove

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

True

Solution steps

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

Manipulating left side

1 - cos (x) sin (x) cot (x)

Express with sin, cos

1 - cos (x) cot (x) sin (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin (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)

= 1 - cos^{2} (x)

= 1 - cos^{2} (x)

= 1 - cos^{2} (x)

Rewrite using trig identities

1 - cos^{2} (x)

Use the Pythagorean identity:

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

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

= sin^{2} (x)

= sin^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

cot (x) + tan (x) = sec (x) \cdot csc (x)

cot (4 x) = \frac{1 - tan ^{2} ( 2 x )}{2 tan ( 2 x )}

sin (16 x) = 2 sin (8 x) cos (8 x)

- cos (2 θ) = \frac{1 - cot ^{2} ( θ )}{1 + cot ^{2} ( θ )}

sin (\frac{π}{2} - x) sec (x) = 1

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