Is tan^2(y)(csc^2(y)-1)=1 ?

The answer to whether tan^2(y)(csc^2(y)-1)=1 is True

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prove tan^2(y)(csc^2(y)-1)=1

prove

tan^{2} (y) (csc^{2} (y) - 1) = 1

True

Solution steps

tan^{2} (y) (csc^{2} (y) - 1) = 1

Manipulating left side

tan^{2} (y) (csc^{2} (y) - 1)

Express with sin, cos

(- 1 + csc^{2} (y)) tan^{2} (y)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

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

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Join

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

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

Convert element to fraction:

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

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

Multiply:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (y)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

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

1 - cos (θ) = sin (θ)

cos (2 π + x) = cos (x)

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

sin (x) = cos (x)

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