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

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

p ro v e cos (θ) = - \frac{3}{4}

p ro v e sin (75) = sin (\frac{150}{2})

p ro v e sec (θ) + 1 = \frac{cos ( θ ) + 1}{cos ( θ )}

p ro v e 9 sin (7 π - x) = 9 sin (x)

p ro v e \frac{cot ( x ) + tan ( x )}{cot ( x )} = sec^{2} (x)

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