Is cot^2(x)*sec^2(x)=csc^2(x) ?

The answer to whether cot^2(x)*sec^2(x)=csc^2(x) is True

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prove cot^2(x)*sec^2(x)=csc^2(x)

prove

cot^{2} (x) \cdot sec^{2} (x) = csc^{2} (x)

True

Solution steps

cot^{2} (x) sec^{2} (x) = csc^{2} (x)

Manipulating left side

cot^{2} (x) sec^{2} (x)

Express with sin, cos

cot^{2} (x) sec^{2} (x)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (x)

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{( \frac{1}{c s c ( x )} ) ^{2}}

Simplify

\frac{1}{( \frac{1}{c s c ( x )} ) ^{2}}

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

= \frac{1}{\frac{1}{c s c ^{2} ( x )}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

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

Apply rule

\frac{a}{1} = a

= csc^{2} (x)

csc^{2} (x)

csc^{2} (x)

We showed that the two sides could take the same form

\Rightarrow True

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

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

cos (π - t) = - cos (t)

sin (\frac{θ}{2}) = 12 - 20

\frac{5 + 5 cos ( 2 x )}{sin ( 2 x )} = 5 cot (x)

Is cot^2(x)*sec^2(x)=csc^2(x) ?
The answer to whether cot^2(x)*sec^2(x)=csc^2(x) is True