Is tan(θ)cot(θ)=cos^2(θ)+sin^2(θ) ?

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

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prove tan(θ)cot(θ)=cos^2(θ)+sin^2(θ)

prove

tan (θ) cot (θ) = cos^{2} (θ) + sin^{2} (θ)

True

Solution steps

tan (θ) cot (θ) = cos^{2} (θ) + sin^{2} (θ)

Manipulating left side

tan (θ) cot (θ)

Express with sin, cos

cot (θ) tan (θ)

Use the basic trigonometric identity:

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

= \frac{cos ( θ )}{sin ( θ )} tan (θ)

Use the basic trigonometric identity:

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

= \frac{cos ( θ )}{sin ( θ )} \cdot \frac{sin ( θ )}{cos ( θ )}

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= \frac{sin ( θ )}{sin ( θ )}

Cancel the common factor:

sin (θ)

= 1

= 1

= 1

Rewrite using trig identities

1

Use the Pythagorean identity:

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

= cos^{2} (θ) + sin^{2} (θ)

= cos^{2} (θ) + sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

cos (θ) = sin (3 θ + 62)

sin (π \cdot x) + sin (π \cdot (10 - x)) = 0

cot (2 θ) = \frac{1}{2} (tan (θ) - cot (θ))

sin (4 x) = (2) (sin (2 x)) (cos (2 x))

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

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