Is sin^2(θ)(1+cot^2(θ))=1 ?

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

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prove sin^2(θ)(1+cot^2(θ))=1

prove

True

Solution steps

sin^{2} (θ) (1 + cot^{2} (θ)) = 1

Manipulating left side

sin^{2} (θ) (1 + cot^{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

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

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