Is cos(θ)+cos(θ)tan^2(θ)=sec(θ) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

prove cos(θ)+cos(θ)tan^2(θ)=sec(θ)

prove

cos (θ) + cos (θ) tan^{2} (θ) = sec (θ)

True

Solution steps

cos (θ) + cos (θ) tan^{2} (θ) = sec (θ)

Manipulating left side

cos (θ) + cos (θ) tan^{2} (θ)

Express with sin, cos

cos (θ) + cos (θ) tan^{2} (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

(\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{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 )}

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

Convert element to fraction:

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

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

cos (θ) cos (θ) + sin^{2} (θ) = cos^{2} (θ) + sin^{2} (θ)

cos (θ) cos (θ) + sin^{2} (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)

= cos^{2} (θ) + sin^{2} (θ)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= \frac{1}{cos ( θ )}

= \frac{1}{cos ( θ )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( θ )}}

Simplify

\frac{1}{\frac{1}{s e c ( θ )}}

Apply the fraction rule:

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

= \frac{sec ( θ )}{1}

Apply rule

\frac{a}{1} = a

= sec (θ)

sec (θ)

sec (θ)

We showed that the two sides could take the same form

\Rightarrow True

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