Is ((tan^2(t)))/(sec(t))=sec(t)-cos(t) ?

The answer to whether ((tan^2(t)))/(sec(t))=sec(t)-cos(t) is True

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Popular Trigonometry >

prove ((tan^2(t)))/(sec(t))=sec(t)-cos(t)

Solution

prove

\frac{( tan ^{2} ( t ) )}{sec ( t )} = sec (t) - cos (t)

Solution

True

Solution steps

\frac{tan ^{2} ( t )}{sec ( t )} = sec (t) - cos (t)

Manipulating left side

\frac{tan ^{2} ( t )}{sec ( t )}

Express with sin, cos

\frac{tan ^{2} ( t )}{sec ( t )}

Use the basic trigonometric identity:

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

= \frac{( \frac{s i n ( t )}{c o s ( t )} ) ^{2}}{sec ( t )}

Use the basic trigonometric identity:

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

= \frac{( \frac{s i n ( t )}{c o s ( t )} ) ^{2}}{\frac{1}{c o s ( t )}}

Simplify

\frac{( \frac{s i n ( t )}{c o s ( t )} ) ^{2}}{\frac{1}{c o s ( t )}} : \frac{sin ^{2} ( t )}{cos ( t )}

\frac{( \frac{s i n ( t )}{c o s ( t )} ) ^{2}}{\frac{1}{c o s ( t )}}

Apply the fraction rule:

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

= \frac{( \frac{s i n ( t )}{c o s ( t )} ) ^{2} cos ( t )}{1}

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

(\frac{sin ( t )}{cos ( t )})^{2}

Apply exponent rule:

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

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

= \frac{\frac{s i n ^{2} ( t )}{c o s ^{2} ( t )} cos ( t )}{1}

Multiply

\frac{sin ^{2} ( t )}{cos ^{2} ( t )} cos (t) : \frac{sin ^{2} ( t )}{cos ( t )}

\frac{sin ^{2} ( t )}{cos ^{2} ( t )} cos (t)

Multiply fractions:

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

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

Cancel the common factor:

cos (t)

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

= \frac{\frac{s i n ^{2} ( t )}{c o s ( t )}}{1}

Apply the fraction rule:

\frac{a}{1} = a

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

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

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

Rewrite using trig identities

\frac{sin ^{2} ( t )}{cos ( t )}

Use the Pythagorean identity:

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

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

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

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

Manipulating right side

sec (t) - cos (t)

Express with sin, cos

- cos (t) + sec (t)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (t) cos (t) + 1

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

cos (t) cos (t)

Apply exponent rule:

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

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

= cos^{1 + 1} (t)

Add the numbers:

1 + 1 = 2

= cos^{2} (t)

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

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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