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) - cos^{2} (θ)

Express with sin, cos

- cos^{2} (θ) + cot (θ) tan (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

- cos^{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

= - cos^{2} (θ) + 1

= - cos^{2} (θ) + 1

= 1 - cos^{2} (θ)

Rewrite using trig identities

1 - cos^{2} (θ)

Use the Pythagorean identity:

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

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

= sin^{2} (θ)

= sin^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

csc^{2} (- θ) - 1 = cot^{2} (θ)

\frac{( sec ( θ ) - tan ( θ ) ) ^{2} + 1}{sec ( θ ) csc ( θ ) - tan ( θ ) csc ( θ )} = 2 tan (θ)

\frac{1}{csc ( x ) - cot ( x )} = csc (x) + cot (x)

(sec (x) + tan (x)) (sec (x) - tan (x)) = 1

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