Is cot^2(y)(sec^2(y)-1)=1 ?

The answer to whether cot^2(y)(sec^2(y)-1)=1 is True

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prove cot^2(y)(sec^2(y)-1)=1

prove

cot^{2} (y) (sec^{2} (y) - 1) = 1

True

Solution steps

cot^{2} (y) (sec^{2} (y) - 1) = 1

Manipulating left side

cot^{2} (y) (sec^{2} (y) - 1)

Express with sin, cos

(- 1 + sec^{2} (y)) cot^{2} (y)

Use the basic trigonometric identity:

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

= (- 1 + (\frac{1}{cos ( y )})^{2}) cot^{2} (y)

Use the basic trigonometric identity:

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

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

Simplify

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

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

(\frac{1}{cos ( y )})^{2} = \frac{1}{cos ^{2} ( y )}

(\frac{1}{cos ( y )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( y )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( y )}

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

= \frac{cos ^{2} ( y ) ( - 1 + \frac{1}{c o s ^{2} ( y )} )}{sin ^{2} ( y )}

Join

- 1 + \frac{1}{cos ^{2} ( y )} : \frac{- cos ^{2} ( y ) + 1}{cos ^{2} ( y )}

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

Convert element to fraction:

1 = \frac{1 cos ^{2} ( y )}{cos ^{2} ( y )}

= - \frac{1 \cdot cos ^{2} ( y )}{cos ^{2} ( y )} + \frac{1}{cos ^{2} ( y )}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot cos ^{2} ( y ) + 1}{cos ^{2} ( y )}

Multiply:

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

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

= \frac{\frac{- c o s ^{2} ( y ) + 1}{c o s ^{2} ( y )} cos ^{2} ( y )}{sin ^{2} ( y )}

Multiply

cos^{2} (y) \frac{- cos ^{2} ( y ) + 1}{cos ^{2} ( y )} : - cos^{2} (y) + 1

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (y)

= - - cos^{2} (y) + 1

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

= \frac{sin ^{2} ( y )}{sin ^{2} ( y )}

Apply rule

\frac{a}{a} = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

p ro v e tan (x - \frac{π}{2}) = - cot (x)

p ro v e sin (π - θ) = sin (θ)

p ro v e tan (\frac{x}{2}) = csc (x) - cot (x)

p ro v e tan (x) cot (x) = 1

p ro v e \frac{1}{2} (cot (x) + tan (x)) = csc (2 x)

Is cot^2(y)(sec^2(y)-1)=1 ?
The answer to whether cot^2(y)(sec^2(y)-1)=1 is True