Is cot^2(x)=((cos^2(x)))/(1-cos^2(x)) ?

The answer to whether cot^2(x)=((cos^2(x)))/(1-cos^2(x)) is True

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prove cot^2(x)=((cos^2(x)))/(1-cos^2(x))

prove

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

True

Solution steps

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

Manipulating left side

cot^{2} (x)

Express with sin, cos

cot^{2} (x)

Use the basic trigonometric identity:

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

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

Simplify

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

(\frac{cos ( x )}{sin ( x )})^{2}

Apply exponent rule:

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

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

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

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

Rewrite using trig identities

\frac{cos ^{2} ( x )}{sin ^{2} ( x )}

Use the Pythagorean identity:

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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

sin (A) cos (A) tan (A) = sin^{2} (A)

sin (\frac{4 π}{3}) = 3 cos (\frac{2 π}{3})

\frac{sin ( 2 x ) + sin ( 6 x )}{cos ( 2 x ) - cos ( 6 x )} = cot (2 x)

sin^{2} (x) + sin^{2} (θ) = 1

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