Is tan(θ)cot(θ)-sin^2(θ)=cos^2(θ) ?

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

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prove tan(θ)cot(θ)-sin^2(θ)=cos^2(θ)

prove

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

True

Solution steps

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

Manipulating left side

tan (θ) cot (θ) - sin^{2} (θ)

Express with sin, cos

- sin^{2} (θ) + cot (θ) tan (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

- sin^{2} (θ) + \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

= - sin^{2} (θ) + 1

= - sin^{2} (θ) + 1

= 1 - sin^{2} (θ)

Rewrite using trig identities

1 - sin^{2} (θ)

Use the Pythagorean identity:

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

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

= cos^{2} (θ)

= cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

sin (2 x) = \frac{2 cot ( x )}{1 + cot ^{2} ( x )}

sec (θ) = \frac{1}{cos ( θ )}

csc (2 θ) = \frac{csc ( θ )}{2 cos ( θ )}

sin (- x) + csc (x) = cot (x) cos (x)

sin (x + y) = sin (x) cos (y) + cos (x) sin (y)

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