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

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

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

prove

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

True

Solution steps

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

Manipulating left side

sec (x) - \frac{tan ^{2} ( x )}{sec ( x )}

Express with sin, cos

- \frac{tan ^{2} ( x )}{sec ( x )} + sec (x)

Use the basic trigonometric identity:

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

= - \frac{( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{sec ( x )} + sec (x)

Use the basic trigonometric identity:

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

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

Simplify

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

- \frac{( \frac{s i n ( x )}{c o s ( x )} ) ^{2}}{\frac{1}{c o s ( x )}} + \frac{1}{cos ( x )}

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

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

Apply the fraction rule:

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

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

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

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

Apply exponent rule:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

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

Apply the fraction rule:

\frac{a}{1} = a

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

\frac{1 - sin ^{2} ( x )}{cos ( x )}

Use the Pythagorean identity:

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

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

= \frac{cos ^{2} ( x )}{cos ( x )}

Cancel the common factor:

cos (x)

= cos (x)

= cos (x)

We showed that the two sides could take the same form

\Rightarrow True

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

\frac{1}{cos ^{3} ( x )} = sec^{3} (x)

sin (x) + cot (x) (cos (x)) = csc (x)

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

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

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