Is 1+cot^2(θ)= 1/(sin^2(θ)) ?

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

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prove 1+cot^2(θ)= 1/(sin^2(θ))

prove

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

True

Solution steps

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

Manipulating left side

1 + cot^{2} (θ)

Express with sin, cos

1 + cot^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

1 + (\frac{cos ( θ )}{sin ( θ )})^{2}

Apply exponent rule:

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

= 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 )}

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= \frac{1}{sin ^{2} ( θ )}

= \frac{1}{sin ^{2} ( θ )}

We showed that the two sides could take the same form

\Rightarrow True

sin (2 π - x) = - sin (x)

csc (θ) - \frac{sin ( θ )}{1 + cos ( θ )} = cot (θ)

\frac{1}{sin ( x ) cos ( x )} - cot (x) = tan (x)

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

sin^{2} (a) + sin^{2} (a) tan^{2} (a) = tan^{2} (a)

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