Is 4cot^2(y)(sec^2(y)-1)=4 ?

The answer to whether 4cot^2(y)(sec^2(y)-1)=4 is True

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prove 4cot^2(y)(sec^2(y)-1)=4

prove

4 cot^{2} (y) (sec^{2} (y) - 1) = 4

True

Solution steps

4 cot^{2} (y) (sec^{2} (y) - 1) = 4

Manipulating left side

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

Express with sin, cos

(- 1 + sec^{2} (y)) \cdot 4 cot^{2} (y)

Use the basic trigonometric identity:

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

= (- 1 + (\frac{1}{cos ( y )})^{2}) \cdot 4 cot^{2} (y)

Use the basic trigonometric identity:

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

= (- 1 + (\frac{1}{cos ( y )})^{2}) \cdot 4 (\frac{cos ( y )}{sin ( y )})^{2}

Simplify

(- 1 + (\frac{1}{cos ( y )})^{2}) \cdot 4 (\frac{cos ( y )}{sin ( y )})^{2} : \frac{4 ( - cos ^{2} ( y ) + 1 )}{sin ^{2} ( y )}

(- 1 + (\frac{1}{cos ( y )})^{2}) \cdot 4 (\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 )}

= 4 (\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 )}

= 4 \cdot \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 )} ) \cdot 4}{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{4 \cdot \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 )} \cdot 4 : 4 (- cos^{2} (y) + 1)

cos^{2} (y) \frac{- cos ^{2} ( y ) + 1}{cos ^{2} ( y )} \cdot 4

Multiply fractions:

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

= \frac{( - cos ^{2} ( y ) + 1 ) cos ^{2} ( y ) \cdot 4}{cos ^{2} ( y )}

Cancel the common factor:

cos^{2} (y)

= (- cos^{2} (y) + 1) \cdot 4

= \frac{4 ( - cos ^{2} ( y ) + 1 )}{sin ^{2} ( y )}

= \frac{4 ( - cos ^{2} ( y ) + 1 )}{sin ^{2} ( y )}

= \frac{( 1 - cos ^{2} ( y ) ) \cdot 4}{sin ^{2} ( y )}

Rewrite using trig identities

\frac{( 1 - cos ^{2} ( y ) ) \cdot 4}{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 ) \cdot 4}{sin ^{2} ( y )}

Cancel the common factor:

sin^{2} (y)

= 4

= 4

We showed that the two sides could take the same form

\Rightarrow True

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

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

cot^{2} (x) (1 + tan^{2} (x)) = csc^{2} (x)

\frac{sec ^{2} ( a ) cot ( a )}{csc ^{2} ( a )} = tan (a)

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

Is 4cot^2(y)(sec^2(y)-1)=4 ?
The answer to whether 4cot^2(y)(sec^2(y)-1)=4 is True