Is (cos(θ))/(1+sin(θ))+tan(θ)=sec(θ) ?

The answer to whether (cos(θ))/(1+sin(θ))+tan(θ)=sec(θ) is True

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prove (cos(θ))/(1+sin(θ))+tan(θ)=sec(θ)

Solution

prove

\frac{cos ( θ )}{1 + sin ( θ )} + tan (θ) = sec (θ)

Solution

True

Solution steps

\frac{cos ( θ )}{1 + sin ( θ )} + tan (θ) = sec (θ)

Manipulating left side

\frac{cos ( θ )}{1 + sin ( θ )} + tan (θ)

Express with sin, cos

\frac{cos ( θ )}{1 + sin ( θ )} + tan (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

\frac{cos ( θ )}{1 + sin ( θ )} + \frac{sin ( θ )}{cos ( θ )}

Least Common Multiplier of

1 + sin (θ), cos (θ) : cos (θ) (sin (θ) + 1)

1 + sin (θ), cos (θ)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

1 + sin (θ)

cos (θ)

= cos (θ) (sin (θ) + 1)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

cos (θ) (sin (θ) + 1)

For

\frac{cos ( θ )}{1 + sin ( θ )} :

multiply the denominator and numerator by

cos (θ)

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

For

\frac{sin ( θ )}{cos ( θ )} :

multiply the denominator and numerator by

sin (θ) + 1

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

1 - sin^{2} (θ) + (1 + sin (θ)) sin (θ) : sin (θ) + 1

1 - sin^{2} (θ) + (1 + sin (θ)) sin (θ)

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

Expand

sin (θ) (1 + sin (θ)) : sin (θ) + sin^{2} (θ)

sin (θ) (1 + sin (θ))

Apply the distributive law:

a (b + c) = ab + a c

a = sin (θ), b = 1, c = sin (θ)

= sin (θ) \cdot 1 + sin (θ) sin (θ)

= 1 \cdot sin (θ) + sin (θ) sin (θ)

Simplify

1 \cdot sin (θ) + sin (θ) sin (θ) : sin (θ) + sin^{2} (θ)

1 \cdot sin (θ) + sin (θ) sin (θ)

1 \cdot sin (θ) = sin (θ)

1 \cdot sin (θ)

Multiply:

1 \cdot sin (θ) = sin (θ)

= sin (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)

= sin (θ) + sin^{2} (θ)

= sin (θ) + sin^{2} (θ)

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

Simplify

1 - sin^{2} (θ) + sin (θ) + sin^{2} (θ) : sin (θ) + 1

1 - sin^{2} (θ) + sin (θ) + sin^{2} (θ)

Group like terms

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

Add similar elements:

- sin^{2} (θ) + sin^{2} (θ) = 0

= sin (θ) + 1

= sin (θ) + 1

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

Cancel the common factor:

sin (θ) + 1

= \frac{1}{cos ( θ )}

= \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

Popular Examples

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cos (x) \cdot tan (x) = sin (x)

prove (sin(2θ))/(sin(θ))-(cos(2θ))/(cos(θ))=sec(θ)prove

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

prove (sec(x))/(csc(x))+(sin(x))/(cos(x))=2tan(x)prove

\frac{sec ( x )}{csc ( x )} + \frac{sin ( x )}{cos ( x )} = 2 tan (x)

prove (sin(θ)+tan(θ))/(1+sec(θ))=sin(θ)prove

\frac{sin ( θ ) + tan ( θ )}{1 + sec ( θ )} = sin (θ)

prove sec(2θ)=(sec^2(θ))/(1-tan^2(θ))prove

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

Frequently Asked Questions (FAQ)

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