Is (cos(x))/(sin^2(x))=cot(x)csc(x) ?

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

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prove (cos(x))/(sin^2(x))=cot(x)csc(x)

prove

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

True

Solution steps

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

Manipulating right side

cot (x) csc (x)

Express with sin, cos

cot (x) csc (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply:

cos (x) \cdot 1 = cos (x)

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

sin (x) sin (x) = sin^{2} (x)

sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

2 sin (9 π - x) = 2 sin (x)

1 csc (θ) - sin (θ) = cot (θ) cos (θ)

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

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

sin (x) cos (x) tan (x) cot (x) = 1

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