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

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

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

prove

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

True

Solution steps

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

Manipulating left side

cos (u) + tan (u) sin (u)

Express with sin, cos

cos (u) + sin (u) tan (u)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Multiply fractions:

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

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

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

sin (u) sin (u)

Apply exponent rule:

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

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

= sin^{1 + 1} (u)

Add the numbers:

1 + 1 = 2

= sin^{2} (u)

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

cos (u) cos (u)

Apply exponent rule:

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

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

= cos^{1 + 1} (u)

Add the numbers:

1 + 1 = 2

= cos^{2} (u)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec (u)

sec (u)

sec (u)

We showed that the two sides could take the same form

\Rightarrow True

sec (- θ) = sec (θ)

sin (\frac{11 π}{12}) = sin (\frac{3 π}{4} + \frac{π}{6})

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

tan (u) + cot (u) = 2 csc (2 u)

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

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