Is tan^2(θ)(csc^2(θ)-1)=1 ?

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

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prove tan^2(θ)(csc^2(θ)-1)=1

prove

tan^{2} (θ) (csc^{2} (θ) - 1) = 1

True

Solution steps

tan^{2} (θ) (csc^{2} (θ) - 1) = 1

Manipulating left side

tan^{2} (θ) (csc^{2} (θ) - 1)

Express with sin, cos

(- 1 + csc^{2} (θ)) tan^{2} (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

(\frac{1}{sin ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sin ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

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

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

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

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

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

Multiply fractions:

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

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

Join

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

- 1 + \frac{1}{sin ^{2} ( θ )}

Convert element to fraction:

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

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

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} ( θ ) + 1}{sin ^{2} ( θ )}

Multiply:

1 \cdot sin^{2} (θ) = sin^{2} (θ)

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (θ)

= - - sin^{2} (θ) + 1

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

cos (x) (tan (x) + 1) = sin (x) + cos (x)

tan^{2} (x) + sec^{2} (x) = 1 + 2 tan^{2} (x)

\frac{sin ( 2 a )}{1 + cos ( 2 a )} = tan (a)

tan (x) = tan (x + π)

1 + csc (a) = csc (a) (1 + sin (a))

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