Is (cos(a))/(1-sin(a))-tan(a)=sec(a) ?

The answer to whether (cos(a))/(1-sin(a))-tan(a)=sec(a) is True

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Popular Trigonometry >

prove (cos(a))/(1-sin(a))-tan(a)=sec(a)

Solution

prove

\frac{cos ( a )}{1 - sin ( a )} - tan (a) = sec (a)

Solution

True

Solution steps

\frac{cos ( a )}{1 - sin ( a )} - tan (a) = sec (a)

Manipulating left side

\frac{cos ( a )}{1 - sin ( a )} - tan (a)

Express with sin, cos

\frac{cos ( a )}{1 - sin ( a )} - tan (a)

Use the basic trigonometric identity:

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

= \frac{cos ( a )}{1 - sin ( a )} - \frac{sin ( a )}{cos ( a )}

Simplify

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

\frac{cos ( a )}{1 - sin ( a )} - \frac{sin ( a )}{cos ( a )}

Least Common Multiplier of

1 - sin (a), cos (a) : cos (a) (- sin (a) + 1)

1 - sin (a), cos (a)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

1 - sin (a)

cos (a)

= cos (a) (- sin (a) + 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 (a) (- sin (a) + 1)

For

\frac{cos ( a )}{1 - sin ( a )} :

multiply the denominator and numerator by

cos (a)

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

For

\frac{sin ( a )}{cos ( a )} :

multiply the denominator and numerator by

- sin (a) + 1

\frac{sin ( a )}{cos ( a )} = \frac{sin ( a ) ( - sin ( a ) + 1 )}{cos ( a ) ( - sin ( a ) + 1 )}

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

Since the denominators are equal, combine the fractions:

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

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

Expand

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

1 - sin^{2} (a) - (1 - sin (a)) sin (a)

= 1 - sin^{2} (a) - sin (a) (1 - sin (a))

Expand

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

- sin (a) (1 - sin (a))

Apply the distributive law:

a (b - c) = ab - a c

a = - sin (a), b = 1, c = sin (a)

= - sin (a) \cdot 1 - (- sin (a)) sin (a)

Apply minus-plus rules

- (- a) = a

= - 1 \cdot sin (a) + sin (a) sin (a)

Simplify

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

- 1 \cdot sin (a) + sin (a) sin (a)

1 \cdot sin (a) = sin (a)

1 \cdot sin (a)

Multiply:

1 \cdot sin (a) = sin (a)

= sin (a)

sin (a) sin (a) = sin^{2} (a)

sin (a) sin (a)

Apply exponent rule:

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

sin (a) sin (a) = sin^{1 + 1} (a)

= sin^{1 + 1} (a)

Add the numbers:

1 + 1 = 2

= sin^{2} (a)

= - sin (a) + sin^{2} (a)

= - sin (a) + sin^{2} (a)

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

Simplify

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

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

Group like terms

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

Add similar elements:

- sin^{2} (a) + sin^{2} (a) = 0

= - sin (a) + 1

= - sin (a) + 1

= \frac{- sin ( a ) + 1}{cos ( a ) ( - sin ( a ) + 1 )}

Cancel the common factor:

- sin (a) + 1

= \frac{1}{cos ( a )}

= \frac{1}{cos ( a )}

= \frac{1}{cos ( a )}

Rewrite using trig identities

Use the basic trigonometric identity:

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

\frac{1}{\frac{1}{s e c ( a )}}

Simplify

\frac{1}{\frac{1}{s e c ( a )}}

Apply the fraction rule:

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

= \frac{sec ( a )}{1}

Apply rule

\frac{a}{1} = a

= sec (a)

sec (a)

sec (a)

We showed that the two sides could take the same form

\Rightarrow True

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