Is tan(-a)=(cos(pi/2+a))/(cos(a)) ?

The answer to whether tan(-a)=(cos(pi/2+a))/(cos(a)) is True

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prove tan(-a)=(cos(pi/2+a))/(cos(a))

prove

tan (- a) = \frac{cos ( \frac{π}{2} + a )}{cos ( a )}

True

Solution steps

tan (- a) = \frac{cos ( \frac{π}{2} + a )}{cos ( a )}

Manipulating left side

tan (- a)

Use the negative angle identity:

tan (- x) = - tan (x)

= - tan (a)

Express with sin, cos

- tan (a)

Use the basic trigonometric identity:

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

= - \frac{sin ( a )}{cos ( a )}

= - \frac{sin ( a )}{cos ( a )}

Manipulating right side

\frac{cos ( \frac{π}{2} + a )}{cos ( a )}

Rewrite using trig identities

cos (\frac{π}{2} + a)

Use the Angle Sum identity:

cos (s + t) = cos (s) cos (t) - sin (s) sin (t)

= cos (\frac{π}{2}) cos (a) - sin (\frac{π}{2}) sin (a)

Simplify

cos (\frac{π}{2}) cos (a) - sin (\frac{π}{2}) sin (a) : - sin (a)

cos (\frac{π}{2}) cos (a) - sin (\frac{π}{2}) sin (a)

cos (\frac{π}{2}) cos (a) = 0

cos (\frac{π}{2}) cos (a)

Simplify

cos (\frac{π}{2}) : 0

cos (\frac{π}{2})

Use the following trivial identity:

cos (\frac{π}{2}) = 0

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= 0

= 0 \cdot cos (a)

Apply rule

0 \cdot a = 0

= 0

sin (\frac{π}{2}) sin (a) = sin (a)

sin (\frac{π}{2}) sin (a)

Simplify

sin (\frac{π}{2}) : 1

sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 1

= 1 \cdot sin (a)

Multiply:

1 \cdot sin (a) = sin (a)

= sin (a)

= 0 - sin (a)

0 - sin (a) = - sin (a)

= - sin (a)

= - sin (a)

= \frac{- sin ( a )}{cos ( a )}

Apply the fraction rule:

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

= - \frac{sin ( a )}{cos ( a )}

We showed that the two sides could take the same form

\Rightarrow True

(1 - cos^{2} (x)) (cot^{2} (x) + 1) = 1

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

cos^{8} (x) = sin^{7} (x) cos^{1} (x)

(cot (x) + 1)^{2} - 2 cot (x) = csc^{2} (x)

csc (x) (2) = sin (x) (\frac{1}{2})

Is tan(-a)=(cos(pi/2+a))/(cos(a)) ?
The answer to whether tan(-a)=(cos(pi/2+a))/(cos(a)) is True