Is (sec^2(θ))/(sec^2(θ)-1)=csc^2(θ) ?

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

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Popular Trigonometry >

prove (sec^2(θ))/(sec^2(θ)-1)=csc^2(θ)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

= \frac{( \frac{1}{c o s ( θ )} ) ^{2}}{- 1 + ( \frac{1}{c o s ( θ )} ) ^{2}}

Simplify

\frac{( \frac{1}{c o s ( θ )} ) ^{2}}{- 1 + ( \frac{1}{c o s ( θ )} ) ^{2}} : \frac{1}{- cos ^{2} ( θ ) + 1}

\frac{( \frac{1}{c o s ( θ )} ) ^{2}}{- 1 + ( \frac{1}{c o s ( θ )} ) ^{2}}

(\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{( \frac{1}{c o s ( θ )} ) ^{2}}{- 1 + \frac{1}{c o s ^{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{\frac{1}{c o s ^{2} ( θ )}}{- 1 + \frac{1}{c o s ^{2} ( θ )}}

Apply the fraction rule:

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

= \frac{1}{cos ^{2} ( θ ) ( - 1 + \frac{1}{c o s ^{2} ( θ )} )}

Join

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

- 1 + \frac{1}{cos ^{2} ( θ )}

Convert element to fraction:

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

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

Multiply:

1 \cdot cos^{2} (θ) = cos^{2} (θ)

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

= \frac{1}{\frac{- c o s ^{2} ( θ ) + 1}{c o s ^{2} ( θ )} cos ^{2} ( θ )}

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (θ)

= - - cos^{2} (θ) + 1

= \frac{1}{- cos ^{2} ( θ ) + 1}

= \frac{1}{- cos ^{2} ( θ ) + 1}

= \frac{1}{1 - cos ^{2} ( θ )}

Rewrite using trig identities

\frac{1}{1 - cos ^{2} ( θ )}

Use the Pythagorean identity:

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

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

= \frac{1}{sin ^{2} ( θ )}

= \frac{1}{sin ^{2} ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

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

(\frac{1}{csc ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{csc ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{csc ^{2} ( θ )}

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

Apply the fraction rule:

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

= \frac{csc ^{2} ( θ )}{1}

Apply rule

\frac{a}{1} = a

= csc^{2} (θ)

csc^{2} (θ)

csc^{2} (θ)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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