Is sec(θ)=(tan(θ))/(csc(θ))+cos(θ) ?

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

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prove sec(θ)=(tan(θ))/(csc(θ))+cos(θ)

Solution

prove

sec (θ) = \frac{tan ( θ )}{csc ( θ )} + cos (θ)

Solution

True

Solution steps

sec (θ) = \frac{tan ( θ )}{csc ( θ )} + cos (θ)

Manipulating right side

\frac{tan ( θ )}{csc ( θ )} + cos (θ)

Express with sin, cos

cos (θ) + \frac{tan ( θ )}{csc ( θ )}

Use the basic trigonometric identity:

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

= cos (θ) + \frac{\frac{s i n ( θ )}{c o s ( θ )}}{csc ( θ )}

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Refine

= \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

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Frequently Asked Questions (FAQ)

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