Is cot(u)tan(u)-sin^2(u)=cos^2(u) ?

The answer to whether cot(u)tan(u)-sin^2(u)=cos^2(u) is True

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prove cot(u)tan(u)-sin^2(u)=cos^2(u)

prove

cot (u) tan (u) - sin^{2} (u) = cos^{2} (u)

True

Solution steps

cot (u) tan (u) - sin^{2} (u) = cos^{2} (u)

Manipulating left side

cot (u) tan (u) - sin^{2} (u)

Express with sin, cos

- sin^{2} (u) + cot (u) tan (u)

Use the basic trigonometric identity:

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

= - sin^{2} (u) + \frac{cos ( u )}{sin ( u )} tan (u)

Use the basic trigonometric identity:

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

= - sin^{2} (u) + \frac{cos ( u )}{sin ( u )} \cdot \frac{sin ( u )}{cos ( u )}

Simplify

- sin^{2} (u) + \frac{cos ( u )}{sin ( u )} \cdot \frac{sin ( u )}{cos ( u )} : - sin^{2} (u) + 1

- sin^{2} (u) + \frac{cos ( u )}{sin ( u )} \cdot \frac{sin ( u )}{cos ( u )}

\frac{cos ( u )}{sin ( u )} \cdot \frac{sin ( u )}{cos ( u )} = 1

\frac{cos ( u )}{sin ( u )} \cdot \frac{sin ( u )}{cos ( u )}

Multiply fractions:

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

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

Cancel the common factor:

cos (u)

= \frac{sin ( u )}{sin ( u )}

Cancel the common factor:

sin (u)

= 1

= - sin^{2} (u) + 1

= - sin^{2} (u) + 1

= 1 - sin^{2} (u)

Rewrite using trig identities

1 - sin^{2} (u)

Use the Pythagorean identity:

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

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

= cos^{2} (u)

= cos^{2} (u)

We showed that the two sides could take the same form

\Rightarrow True

\frac{1}{cos ( θ )} = \frac{tan ( θ )}{sin ( θ )}

sin (x) \cdot tan (x) = sec (x) - cos (x)

(tan (y) + cot (y)) sin (y) cos (y) = 1

- 2 cos^{2} (x) = sin^{4} (x) - cos^{4} (x) - 1

sin (x) \cdot cot (x) = cos (x)

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