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

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

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

prove

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

True

Solution steps

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

Manipulating right side

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

Express with sin, cos

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

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply the fraction rule:

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

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

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

(\frac{sin ( θ )}{cos ( θ )})^{2}

Apply exponent rule:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (θ)

= \frac{sin ^{2} ( θ )}{cos ( θ )}

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

Apply the fraction rule:

\frac{a}{1} = a

= \frac{sin ^{2} ( θ )}{cos ( θ )}

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

Apply rule

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

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

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

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

Rewrite using trig identities

\frac{1 - sin ^{2} ( θ )}{cos ( θ )}

Use the Pythagorean identity:

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

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

= \frac{cos ^{2} ( θ )}{cos ( θ )}

Cancel the common factor:

cos (θ)

= cos (θ)

= cos (θ)

We showed that the two sides could take the same form

\Rightarrow True

csc (x^{2}) sec (x^{2}) = csc (x^{2}) + sec (x^{2})

tan (\frac{θ}{2}) = csc^{2} (θ) - cot^{2} (θ)

4 + 4 cos (2 x) = 8 cos^{2} (x)

\frac{sec ( θ ) csc ( θ )}{2} = csc (2 θ)

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

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