Is sin^2(θ)sec^2(θ)csc^2(θ)=sec^2(θ) ?

The answer to whether sin^2(θ)sec^2(θ)csc^2(θ)=sec^2(θ) is True

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prove sin^2(θ)sec^2(θ)csc^2(θ)=sec^2(θ)

prove

sin^{2} (θ) sec^{2} (θ) csc^{2} (θ) = sec^{2} (θ)

True

Solution steps

sin^{2} (θ) sec^{2} (θ) csc^{2} (θ) = sec^{2} (θ)

Manipulating left side

sin^{2} (θ) sec^{2} (θ) csc^{2} (θ)

Express with sin, cos

csc^{2} (θ) sec^{2} (θ) sin^{2} (θ)

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

(\frac{1}{cos ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{cos ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{cos ^{2} ( θ )}

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

Multiply fractions:

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

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

Cancel the common factor:

sin^{2} (θ)

= \frac{1 \cdot 1}{cos ^{2} ( θ )}

Multiply the numbers:

1 \cdot 1 = 1

= \frac{1}{cos ^{2} ( θ )}

= \frac{1}{cos ^{2} ( θ )}

= \frac{1}{cos ^{2} ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{( \frac{1}{s e c ( θ )} ) ^{2}}

Simplify

\frac{1}{( \frac{1}{s e c ( θ )} ) ^{2}}

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

(\frac{1}{sec ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sec ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sec ^{2} ( θ )}

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

Apply the fraction rule:

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

= \frac{sec ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= sec^{2} (θ)

sec^{2} (θ)

sec^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

\frac{csc ( θ ) - sin ( θ )}{sin ( θ )} = cot^{2} (θ)

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

csc (x) sin (x) = 1

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

\frac{cos ( A )}{1 + sin ( A )} + \frac{cos ( A )}{1 - sin ( A )} = 2 sec (A)

Is sin^2(θ)sec^2(θ)csc^2(θ)=sec^2(θ) ?
The answer to whether sin^2(θ)sec^2(θ)csc^2(θ)=sec^2(θ) is True