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

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

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prove (cos(t))/(sec(t)-tan(t))=1+sin(t)

Solution

prove

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

Solution

True

Solution steps

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

Manipulating left side

\frac{cos ( t )}{sec ( t ) - tan ( t )}

Express with sin, cos

\frac{cos ( t )}{sec ( t ) - tan ( t )}

Use the basic trigonometric identity:

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

= \frac{cos ( t )}{\frac{1}{c o s ( t )} - tan ( t )}

Use the basic trigonometric identity:

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

= \frac{cos ( t )}{\frac{1}{c o s ( t )} - \frac{s i n ( t )}{c o s ( t )}}

Simplify

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

\frac{cos ( t )}{\frac{1}{c o s ( t )} - \frac{s i n ( t )}{c o s ( t )}}

Combine the fractions

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

Apply rule

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

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

= \frac{cos ( t )}{\frac{1 - s i n ( t )}{c o s ( t )}}

Apply the fraction rule:

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

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

cos (t) cos (t) = cos^{2} (t)

cos (t) cos (t)

Apply exponent rule:

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

cos (t) cos (t) = cos^{1 + 1} (t)

= cos^{1 + 1} (t)

Add the numbers:

1 + 1 = 2

= cos^{2} (t)

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Simplify

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

\frac{1 - sin ^{2} ( t )}{1 - sin ( t )}

Factor

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

1 - sin^{2} (t)

Factor out common term

- 1

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

Factor

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

sin^{2} (t) - 1

Rewrite

1

1^{2}

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

= (sin (t) + 1) (sin (t) - 1)

= - (sin (t) + 1) (sin (t) - 1)

= - \frac{( sin ( t ) + 1 ) ( sin ( t ) - 1 )}{1 - sin ( t )}

Cancel

- \frac{( sin ( t ) + 1 ) ( sin ( t ) - 1 )}{1 - sin ( t )} : sin (t) + 1

- \frac{( sin ( t ) + 1 ) ( sin ( t ) - 1 )}{1 - sin ( t )}

- sin (t) + 1 = - (sin (t) - 1)

= - \frac{( sin ( t ) + 1 ) ( sin ( t ) - 1 )}{- ( sin ( t ) - 1 )}

Refine

= \frac{( sin ( t ) + 1 ) ( sin ( t ) - 1 )}{sin ( t ) - 1}

Cancel the common factor:

sin (t) - 1

= sin (t) + 1

= sin (t) + 1

= sin (t) + 1

= sin (t) + 1

= 1 + sin (t)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

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