Is 1/(cot^2(x))+sec(x)cos(x)=csc^2(x) ?

The answer to whether 1/(cot^2(x))+sec(x)cos(x)=csc^2(x) is False

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prove 1/(cot^2(x))+sec(x)cos(x)=csc^2(x)

prove

\frac{1}{cot ^{2} ( x )} + sec (x) cos (x) = csc^{2} (x)

False

Solution steps

\frac{1}{cot ^{2} ( x )} + sec (x) cos (x) = csc^{2} (x)

Show that the two sides are not equal

For

\frac{1}{cot ^{2} ( x )} + sec (x) cos (x) = csc^{2} (x)

plug in

x = 1

\frac{1}{cot ^{2} ( 1 )} + sec (1) cos (1) = 3.42551 \dots

\frac{1}{cot ^{2} ( 1 )} + sec (1) cos (1)

Refine to a decimal form

= 3.42551 \dots

csc^{2} (1) = 1.41228 \dots

csc^{2} (1)

Refine to a decimal form

= 1.41228 \dots

The two sides are not equal

\Rightarrow False

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

1 csc (θ) - (cos (θ)) (cot (θ)) = sin (θ)

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

sin^{2} (3 x) = 9 sin^{3} (x) cos^{3} (x)

8 tan (x) sin (x) - 8 tan (x) = 0

Is 1/(cot^2(x))+sec(x)cos(x)=csc^2(x) ?
The answer to whether 1/(cot^2(x))+sec(x)cos(x)=csc^2(x) is False