Is cot(x)tan(x)-(sin^2(x)+cos^2(x))=0 ?

The answer to whether cot(x)tan(x)-(sin^2(x)+cos^2(x))=0 is True

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prove cot(x)tan(x)-(sin^2(x)+cos^2(x))=0

prove

cot (x) tan (x) - (sin^{2} (x) + cos^{2} (x)) = 0

True

Solution steps

cot (x) tan (x) - (sin^{2} (x) + cos^{2} (x)) = 0

Manipulating left side

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= cot (x) tan (x) - 1

= cot (x) tan (x) - 1

Express with sin, cos

- 1 + cot (x) tan (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

- 1 + \frac{cos ( x )}{sin ( x )} \cdot \frac{sin ( x )}{cos ( x )} = 0

- 1 + \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

= - 1 + 1

Simplify

= 0

= 0

= 0

We showed that the two sides could take the same form

\Rightarrow True

Is cot(x)tan(x)-(sin^2(x)+cos^2(x))=0 ?
The answer to whether cot(x)tan(x)-(sin^2(x)+cos^2(x))=0 is True