Is cos(-x)+(sin(-x))/(cot(-x))=sec(x) ?

The answer to whether cos(-x)+(sin(-x))/(cot(-x))=sec(x) is True

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Popular Trigonometry >

prove cos(-x)+(sin(-x))/(cot(-x))=sec(x)

Solution

prove

cos (- x) + \frac{sin ( - x )}{cot ( - x )} = sec (x)

Solution

True

Solution steps

cos (- x) + \frac{sin ( - x )}{cot ( - x )} = sec (x)

Manipulating left side

cos (- x) + \frac{sin ( - x )}{cot ( - x )}

Use the negative angle identity:

sin (- x) = - sin (x)

= cos (- x) + \frac{- sin ( x )}{cot ( - x )}

Use the negative angle identity:

cos (- x) = cos (x)

= cos (x) + \frac{- sin ( x )}{cot ( - x )}

Use the negative angle identity:

cot (- x) = - cot (x)

= cos (x) + \frac{- sin ( x )}{- cot ( x )}

Express with sin, cos

cos (x) + \frac{- sin ( x )}{- cot ( x )}

Use the basic trigonometric identity:

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

= cos (x) + \frac{- sin ( x )}{- \frac{c o s ( x )}{s i n ( x )}}

Simplify

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

cos (x) + \frac{- sin ( x )}{- \frac{c o s ( x )}{s i n ( x )}}

\frac{- sin ( x )}{- \frac{c o s ( x )}{s i n ( x )}} = \frac{sin ^{2} ( x )}{cos ( x )}

\frac{- sin ( x )}{- \frac{c o s ( x )}{s i n ( x )}}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{sin ( x )}{\frac{c o s ( x )}{s i n ( x )}}

Apply the fraction rule:

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

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

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

sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

cos (x) cos (x)

Apply exponent rule:

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

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

= cos^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= cos^{2} (x)

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

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

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

Rewrite using trig identities

Use the basic trigonometric identity:

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

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

Simplify

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

Apply the fraction rule:

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

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

Apply rule

\frac{a}{1} = a

= sec (x)

sec (x)

sec (x)

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is cos(-x)+(sin(-x))/(cot(-x))=sec(x) ?
The answer to whether cos(-x)+(sin(-x))/(cot(-x))=sec(x) is True