Is (cot^2(t))/(cos(t))=cot(t)csc(t) ?

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

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prove (cot^2(t))/(cos(t))=cot(t)csc(t)

prove

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

True

Solution steps

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

Manipulating left side

\frac{cot ^{2} ( t )}{cos ( t )}

Express with sin, cos

\frac{cot ^{2} ( t )}{cos ( t )}

Use the basic trigonometric identity:

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

= \frac{( \frac{c o s ( t )}{s i n ( t )} ) ^{2}}{cos ( t )}

Simplify

\frac{( \frac{c o s ( t )}{s i n ( t )} ) ^{2}}{cos ( t )} : \frac{cos ( t )}{sin ^{2} ( t )}

\frac{( \frac{c o s ( t )}{s i n ( t )} ) ^{2}}{cos ( t )}

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

(\frac{cos ( t )}{sin ( t )})^{2}

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

= \frac{\frac{c o s ^{2} ( t )}{s i n ^{2} ( t )}}{cos ( t )}

Apply the fraction rule:

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

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

Cancel the common factor:

cos (t)

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

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

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

Manipulating right side

cot (t) csc (t)

Express with sin, cos

cot (t) csc (t)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Multiply fractions:

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

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

Multiply:

cos (t) \cdot 1 = cos (t)

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

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

sin (t) sin (t)

Apply exponent rule:

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

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

= sin^{1 + 1} (t)

Add the numbers:

1 + 1 = 2

= sin^{2} (t)

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

\frac{sin ( x ) + tan ( x )}{1 + sec ( x )} = sec (x)

- csc^{2} (x) = - 1 - (cot (x))^{2}

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

\frac{1}{csc ( θ )} = - cos (θ)

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

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