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

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

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

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

- cos^{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

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

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

= 1 - cos^{2} (u)

Rewrite using trig identities

1 - cos^{2} (u)

Use the Pythagorean identity:

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

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

= sin^{2} (u)

= sin^{2} (u)

We showed that the two sides could take the same form

\Rightarrow True

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

cos (2 θ) = 1 - 2 sin^{2} (θ)

sin (x) = 1 - cos (x)

\frac{cos ( θ )}{cot ( θ )} = sin (θ)

sin (θ) = \frac{tan ( θ )}{sec ( θ )}

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