Is (sec(v)-cos(v))/(sec(v))=sin^2(v) ?

The answer to whether (sec(v)-cos(v))/(sec(v))=sin^2(v) is True

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prove (sec(v)-cos(v))/(sec(v))=sin^2(v)

prove

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

True

Solution steps

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

Manipulating left side

\frac{sec ( v ) - cos ( v )}{sec ( v )}

Express with sin, cos

\frac{- cos ( v ) + sec ( v )}{sec ( v )}

Use the basic trigonometric identity:

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

= \frac{- cos ( v ) + \frac{1}{c o s ( v )}}{\frac{1}{c o s ( v )}}

Simplify

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

\frac{- cos ( v ) + \frac{1}{c o s ( v )}}{\frac{1}{c o s ( v )}}

Apply the fraction rule:

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

= \frac{( - cos ( v ) + \frac{1}{c o s ( v )} ) cos ( v )}{1}

Join

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

- cos (v) + \frac{1}{cos ( v )}

Convert element to fraction:

cos (v) = \frac{cos ( v ) cos ( v )}{cos ( v )}

= - \frac{cos ( v ) cos ( v )}{cos ( v )} + \frac{1}{cos ( v )}

Since the denominators are equal, combine the fractions:

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

= \frac{- cos ( v ) cos ( v ) + 1}{cos ( v )}

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

- cos (v) cos (v) + 1

cos (v) cos (v) = cos^{2} (v)

cos (v) cos (v)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos (v) cos (v) = cos^{1 + 1} (v)

= cos^{1 + 1} (v)

Add the numbers:

1 + 1 = 2

= cos^{2} (v)

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

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

Multiply fractions:

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

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

Cancel the common factor:

cos (v)

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

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

= 1 - cos^{2} (v)

Rewrite using trig identities

1 - cos^{2} (v)

Use the Pythagorean identity:

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

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

= sin^{2} (v)

= sin^{2} (v)

We showed that the two sides could take the same form

\Rightarrow True

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

sec^{2} (θ) - tan (θ) cot (θ) = tan^{2} (θ)

csc^{2} (θ) (1 - cos^{2} (θ)) = tan (42 0^{\circ})

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

tan^{4} (x) = (1 - sec^{2} (x))^{2}

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