Is cos(t)+tan(t)sin(t)=sec(t) ?

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

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prove cos(t)+tan(t)sin(t)=sec(t)

prove

cos (t) + tan (t) sin (t) = sec (t)

True

Solution steps

cos (t) + tan (t) sin (t) = sec (t)

Manipulating left side

cos (t) + tan (t) sin (t)

Express with sin, cos

cos (t) + sin (t) tan (t)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

sin (t) \frac{sin ( t )}{cos ( t )}

Multiply fractions:

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

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

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

sin (t) sin (t)

Apply exponent rule:

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

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

= sin^{1 + 1} (t)

Add the numbers:

1 + 1 = 2

= sin^{2} (t)

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

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

Convert element to fraction:

cos (t) = \frac{cos ( t ) cos ( t )}{cos ( t )}

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

Since the denominators are equal, combine the fractions:

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

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

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

cos (t) cos (t) + sin^{2} (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)

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

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

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

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

Rewrite using trig identities

\frac{cos ^{2} ( t ) + sin ^{2} ( t )}{cos ( t )}

Use the Pythagorean identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec (t)

sec (t)

sec (t)

We showed that the two sides could take the same form

\Rightarrow True

\frac{1 - cos ( x )}{- 1 + \frac{1}{c o s ( x )}} = cos (x)

cos^{4} (x) - sin^{4} (x) = tan^{2} (x)

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

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

tan^{3} (x) = \frac{sin ^{3} ( x )}{cos ^{3} ( x )}

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