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

The answer to whether cot^2(x)-cos^2(x)=cot^2(x)cot(x) is False

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

prove

cot^{2} (x) - cos^{2} (x) = cot^{2} (x) cot (x)

False

Solution steps

cot^{2} (x) - cos^{2} (x) = cot^{2} (x) cot (x)

Show that the two sides are not equal

For

cot^{2} (x) - cos^{2} (x) = cot^{2} (x) cot (x)

plug in

x = 1

cot^{2} (1) - cos^{2} (1) = 0.12035 \dots

cot^{2} (1) - cos^{2} (1)

Refine to a decimal form

= 0.12035 \dots

cot^{2} (1) cot (1) = 0.26472 \dots

cot^{2} (1) cot (1)

Refine to a decimal form

= 0.26472 \dots

The two sides are not equal

\Rightarrow False

sin^{2} (3 2^{\circ}) + cos^{2} (x) = 1

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

cos (θ) = - \frac{3}{4}

sin (75) = sin (\frac{150}{2})

(sec^{2} (y)) = \frac{1}{cos ^{2} ( y )}

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