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

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

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

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{sec ( θ )}{csc ^{2} ( θ )}

Express with sin, cos

\frac{sec ( θ )}{csc ^{2} ( θ )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Apply the fraction rule:

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

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

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

(\frac{1}{sin ( θ )})^{2}

Apply exponent rule:

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

= \frac{1 ^{2}}{sin ^{2} ( θ )}

Apply rule

1^{a} = 1

1^{2} = 1

= \frac{1}{sin ^{2} ( θ )}

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

Multiply

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

cos (θ) \frac{1}{sin ^{2} ( θ )}

Multiply fractions:

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

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

Multiply:

1 \cdot cos (θ) = cos (θ)

= \frac{cos ( θ )}{sin ^{2} ( θ )}

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

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

= \frac{sin ^{2} ( θ )}{cos ( θ )}

= \frac{sin ^{2} ( θ )}{cos ( θ )}

= \frac{sin ^{2} ( θ )}{cos ( θ )}

Rewrite using trig identities

\frac{sin ^{2} ( θ )}{cos ( θ )}

Use the Pythagorean identity:

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

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

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

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

Manipulating right side

sec (θ) - cos (θ)

Express with sin, cos

- cos (θ) + sec (θ)

Use the basic trigonometric identity:

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

= - cos (θ) + \frac{1}{cos ( θ )}

Simplify

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

- cos (θ) + \frac{1}{cos ( θ )}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (θ) cos (θ) + 1

cos (θ) cos (θ) = cos^{2} (θ)

cos (θ) cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos^{2} (θ)

= - cos^{2} (θ) + 1

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

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

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

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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