Is (1-sin^2(θ))/(sin^2(θ))=cot^2(θ) ?

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

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

prove

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

True

Solution steps

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

Manipulating right side

cot^{2} (θ)

Express with sin, cos

cot^{2} (θ)

Use the basic trigonometric identity:

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

= (\frac{cos ( θ )}{sin ( θ )})^{2}

Simplify

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

(\frac{cos ( θ )}{sin ( θ )})^{2}

Apply exponent rule:

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

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

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

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

Rewrite using trig identities

\frac{cos ^{2} ( θ )}{sin ^{2} ( θ )}

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

sin (x) = tan (x) cos (x)

sin (θ) \cdot (cot (θ) + tan (θ)) = sec (θ)

tan (x) + 10 = \frac{sin ( x ) + 10}{cos ( x )}

\frac{( sin ^{2} ( t ) )}{tan ^{2} ( t )} = cos^{2} (t)

sin (A) sec (A) = tan (A)

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