Is ((sec(θ))/(cos(θ)))-((tan(θ))/(cot(θ)))=1 ?

The answer to whether ((sec(θ))/(cos(θ)))-((tan(θ))/(cot(θ)))=1 is True

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

prove ((sec(θ))/(cos(θ)))-((tan(θ))/(cot(θ)))=1

prove

(\frac{sec ( θ )}{cos ( θ )}) - (\frac{tan ( θ )}{cot ( θ )}) = 1

True

Solution steps

\frac{sec ( θ )}{cos ( θ )} - \frac{tan ( θ )}{cot ( θ )} = 1

Manipulating left side

\frac{sec ( θ )}{cos ( θ )} - \frac{tan ( θ )}{cot ( θ )}

Express with sin, cos

\frac{sec ( θ )}{cos ( θ )} - \frac{tan ( θ )}{cot ( θ )}

Use the basic trigonometric identity:

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

= \frac{\frac{1}{c o s ( θ )}}{cos ( θ )} - \frac{tan ( θ )}{cot ( θ )}

Use the basic trigonometric identity:

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

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

\frac{\frac{1}{c o s ( θ )}}{cos ( θ )}

Apply the fraction rule:

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

= \frac{1}{cos ( θ ) cos ( θ )}

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)

= \frac{1}{cos ^{2} ( θ )}

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

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

Divide fractions:

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

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

sin (θ) sin (θ) = sin^{2} (θ)

sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= sin^{2} (θ)

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

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)

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

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

Apply rule

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Apply rule

\frac{a}{a} = 1

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

Is ((sec(θ))/(cos(θ)))-((tan(θ))/(cot(θ)))=1 ?
The answer to whether ((sec(θ))/(cos(θ)))-((tan(θ))/(cot(θ)))=1 is True