Is cot(t)csc(t)=(csc^2(t)-1)/(cos(t)) ?

The answer to whether cot(t)csc(t)=(csc^2(t)-1)/(cos(t)) is True

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prove cot(t)csc(t)=(csc^2(t)-1)/(cos(t))

Solution

prove

cot (t) csc (t) = \frac{csc ^{2} ( t ) - 1}{cos ( t )}

Solution

True

Solution steps

cot (t) csc (t) = \frac{csc ^{2} ( t ) - 1}{cos ( t )}

Manipulating left 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 )}

Manipulating right side

\frac{csc ^{2} ( t ) - 1}{cos ( t )}

Express with sin, cos

\frac{- 1 + csc ^{2} ( t )}{cos ( t )}

Use the basic trigonometric identity:

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

= \frac{- 1 + ( \frac{1}{s i n ( t )} ) ^{2}}{cos ( t )}

Simplify

\frac{- 1 + ( \frac{1}{s i n ( t )} ) ^{2}}{cos ( t )} : \frac{- sin ^{2} ( t ) + 1}{sin ^{2} ( t ) cos ( t )}

\frac{- 1 + ( \frac{1}{s i n ( t )} ) ^{2}}{cos ( t )}

(\frac{1}{sin ( t )})^{2} = \frac{1}{sin ^{2} ( t )}

(\frac{1}{sin ( t )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sin ^{2} ( t )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( t )}

= \frac{- 1 + \frac{1}{s i n ^{2} ( t )}}{cos ( t )}

Join

- 1 + \frac{1}{sin ^{2} ( t )} : \frac{- sin ^{2} ( t ) + 1}{sin ^{2} ( t )}

- 1 + \frac{1}{sin ^{2} ( t )}

Convert element to fraction:

1 = \frac{1 sin ^{2} ( t )}{sin ^{2} ( t )}

= - \frac{1 \cdot sin ^{2} ( t )}{sin ^{2} ( t )} + \frac{1}{sin ^{2} ( t )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot sin ^{2} ( t ) + 1}{sin ^{2} ( t )}

Multiply:

1 \cdot sin^{2} (t) = sin^{2} (t)

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

= \frac{\frac{- s i n ^{2} ( t ) + 1}{s i n ^{2} ( t )}}{cos ( t )}

Apply the fraction rule:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Cancel the common factor:

cos (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

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Frequently Asked Questions (FAQ)

Is cot(t)csc(t)=(csc^2(t)-1)/(cos(t)) ?
The answer to whether cot(t)csc(t)=(csc^2(t)-1)/(cos(t)) is True