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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

Cancel the common factor:

sin (x)

= 1

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

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

= 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

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

cot^{4} (x) + 2 cot^{2} (x) = csc^{4} (x) - 1

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

cot (3) = \frac{cos ( 3 )}{sin ( 3 )}

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

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