Is tan^2(x)csc^2(x)cot^2(x)sin^2(x)=1 ?

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

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prove tan^2(x)csc^2(x)cot^2(x)sin^2(x)=1

prove

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

True

Solution steps

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

Manipulating left side

tan^{2} (x) csc^{2} (x) cot^{2} (x) sin^{2} (x)

Express with sin, cos

cot^{2} (x) csc^{2} (x) sin^{2} (x) tan^{2} (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

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

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

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

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

Apply exponent rule:

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

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

Cancel the common factor:

sin^{2} (x)

= 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

sec (x) cot (x) sin (x) = 1

- sec (x) sin (x) = - tan (x)

\frac{sin ( 3 x )}{sin ( 2 x )} = 2 cos (x) - \frac{1}{2} sec (x)

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

cos (1 5^{\circ}) = cos (4 5^{\circ} - 3 0^{\circ})

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