Is sin(θ)tan(θ)+cos(θ)=sec(θ) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

prove sin(θ)tan(θ)+cos(θ)=sec(θ)

prove

sin (θ) tan (θ) + cos (θ) = sec (θ)

True

Solution steps

sin (θ) tan (θ) + cos (θ) = sec (θ)

Manipulating left side

sin (θ) tan (θ) + cos (θ)

Express with sin, cos

cos (θ) + sin (θ) tan (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

sin (θ) \frac{sin ( θ )}{cos ( θ )}

Multiply fractions:

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

= \frac{sin ( θ ) sin ( 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) + \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

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

cos^{4} (θ) - sin^{4} (θ) = cos (2 θ)

cot^{2} (y) - cos^{2} (y) = cot^{2} (y) cos^{2} (y)

cot^{2} (x) = (csc (x) - 1) (csc (x) + 1)

sin (θ) - sin (θ) cos^{2} (θ) = sin^{3} (θ)

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