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

The answer to whether cos^2(x)csc(x)sec(x)=cot(x) is True

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prove cos^2(x)csc(x)sec(x)=cot(x)

prove

cos^{2} (x) csc (x) sec (x) = cot (x)

True

Solution steps

cos^{2} (x) csc (x) sec (x) = cot (x)

Manipulating left side

cos^{2} (x) csc (x) sec (x)

Express with sin, cos

cos^{2} (x) csc (x) sec (x)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Use the basic trigonometric identity:

sec (x) = \frac{1}{cos ( x )}

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

Simplify

cos^{2} (x) \frac{1}{sin ( x )} \cdot \frac{1}{cos ( x )} : \frac{cos ( x )}{sin ( x )}

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

Multiply fractions:

a \cdot \frac{b}{c} \cdot \frac{d}{e} = \frac{a \cdot b \cdot d}{c \cdot e}

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

Multiply:

1 \cdot 1 \cdot cos^{2} (x) = cos^{2} (x)

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

Cancel the common factor:

cos (x)

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

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

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

Use the basic trigonometric identity:

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

= cot (x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

sin^{2} (θ) - cos^{2} (θ) sin^{2} (θ) = sin^{4} (θ)

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

(1 + csc (θ)) (1 - csc (θ)) = - cot^{2} (θ)

Is cos^2(x)csc(x)sec(x)=cot(x) ?
The answer to whether cos^2(x)csc(x)sec(x)=cot(x) is True