Is ((sec^2(Q)-1))/(sec^2(Q))=sin^2(Q) ?

The answer to whether ((sec^2(Q)-1))/(sec^2(Q))=sin^2(Q) is True

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prove ((sec^2(Q)-1))/(sec^2(Q))=sin^2(Q)

prove

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

True

Solution steps

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

Manipulating left side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

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

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

Apply exponent rule:

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

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

Apply rule

1^{a} = 1

1^{2} = 1

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

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

Apply the fraction rule:

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

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

Join

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

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

Convert element to fraction:

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

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

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} ( Q ) + 1}{cos ^{2} ( Q )}

Multiply:

1 \cdot cos^{2} (Q) = cos^{2} (Q)

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

cos^{2} (Q)

= - - cos^{2} (Q) + 1

= - cos^{2} (Q) + 1

= 1 - cos^{2} (Q)

Rewrite using trig identities

1 - cos^{2} (Q)

Use the Pythagorean identity:

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

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

= sin^{2} (Q)

= sin^{2} (Q)

We showed that the two sides could take the same form

\Rightarrow True

1 - tan^{2} (A) cos^{2} (A) = cos^{2} (A)

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

tan (x) + cot (x) = \frac{1}{sin ( x ) \cdot cos ( x )}

\frac{1 - 4 sec ^{2} ( x )}{1 + 2 sec ( x )} = 1 - 2 sec (x)

sin (\frac{3 π}{4}) = \frac{2}{2}

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