Is (1-cot^2(θ))(sin^2(θ))=1-2cos^2(θ) ?

The answer to whether (1-cot^2(θ))(sin^2(θ))=1-2cos^2(θ) is True

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prove (1-cot^2(θ))(sin^2(θ))=1-2cos^2(θ)

prove

(1 - cot^{2} (θ)) (sin^{2} (θ)) = 1 - 2 cos^{2} (θ)

True

Solution steps

(1 - cot^{2} (θ)) sin^{2} (θ) = 1 - 2 cos^{2} (θ)

Manipulating left side

(1 - cot^{2} (θ)) sin^{2} (θ)

Express with sin, cos

(1 - cot^{2} (θ)) sin^{2} (θ)

Use the basic trigonometric identity:

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

= (1 - (\frac{cos ( θ )}{sin ( θ )})^{2}) sin^{2} (θ)

Simplify

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Join

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

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

Convert element to fraction:

1 = \frac{1 sin ^{2} ( θ )}{sin ^{2} ( θ )}

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

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

Multiply:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (θ)

= sin^{2} (θ) - cos^{2} (θ)

= sin^{2} (θ) - cos^{2} (θ)

= - cos^{2} (θ) + sin^{2} (θ)

Rewrite using trig identities

- cos^{2} (θ) + sin^{2} (θ)

Use the Pythagorean identity:

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

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

= - cos^{2} (θ) + 1 - cos^{2} (θ)

Simplify

- cos^{2} (θ) + 1 - cos^{2} (θ) : - 2 cos^{2} (θ) + 1

- cos^{2} (θ) + 1 - cos^{2} (θ)

Group like terms

= - cos^{2} (θ) - cos^{2} (θ) + 1

Add similar elements:

- cos^{2} (θ) - cos^{2} (θ) = - 2 cos^{2} (θ)

= - 2 cos^{2} (θ) + 1

= - 2 cos^{2} (θ) + 1

= - 2 cos^{2} (θ) + 1

= 1 - 2 cos^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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

(cot (X) - csc (X)) (cot (X) + csc (X)) = - 1

cos (4 x) = cos (3 x + x)

cot (θ) \cdot sec (θ) \cdot sin (θ) = 1

\frac{1}{sec ^{2} ( t )} = (1 + sin (t)) (1 - sin (t))

Is (1-cot^2(θ))(sin^2(θ))=1-2cos^2(θ) ?
The answer to whether (1-cot^2(θ))(sin^2(θ))=1-2cos^2(θ) is True