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

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

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

Solution

prove

tan (- u) \cdot sin (- u) + cos (- u) = sec (u)

Solution

True

Solution steps

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

Manipulating left side

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

Use the negative angle identity:

sin (- x) = - sin (x)

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

Use the negative angle identity:

cos (- x) = cos (x)

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

Use the negative angle identity:

tan (- x) = - tan (x)

= 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 )})

Remove parentheses:

(- a) = - a, - (- a) = a

= 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

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Frequently Asked Questions (FAQ)

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