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

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

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

prove

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

True

Solution steps

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

Manipulating left side

tan^{2} (x) csc^{2} (x) - tan^{2} (x)

Express with sin, cos

- tan^{2} (x) + csc^{2} (x) tan^{2} (x)

Use the basic trigonometric identity:

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

= - (\frac{sin ( x )}{cos ( x )})^{2} + csc^{2} (x) (\frac{sin ( x )}{cos ( x )})^{2}

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

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

(\frac{1}{sin ( x )})^{2} (\frac{sin ( x )}{cos ( x )})^{2}

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

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

Apply exponent rule:

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

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

= \frac{1}{sin ^{2} ( x )} \cdot \frac{sin ^{2} ( x )}{cos ^{2} ( x )}

Multiply fractions:

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

= \frac{1 \cdot sin ^{2} ( x )}{sin ^{2} ( x ) cos ^{2} ( x )}

Cancel the common factor:

sin^{2} (x)

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

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

tan^{2} (x) - (sec (x) + 1) (sec (x) - 1) = 0

csc (u) sin (u) - cos^{2} (u) = sin^{2} (u)

2 cos^{2} (u) - cos (2 u) = 1

sec^{4} (θ) = sec^{2} (θ) tan^{2} (θ) + sec^{2} (θ)

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