Is sec(-θ)=-sec(θ) ?

The answer to whether sec(-θ)=-sec(θ) is False

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prove sec(-θ)=-sec(θ)

prove

sec (- θ) = - sec (θ)

False

Solution steps

sec (- θ) = - sec (θ)

Show that the two sides are not equal

For

sec (- θ) = - sec (θ)

plug in

θ = 0

sec (- 0) = 1

sec (- 0)

Simplify

= sec (0)

Use the following trivial identity:

sec (0) = 1

sec (0)

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

= 1

= 1

- sec (0) = - 1

- sec (0)

Use the following trivial identity:

sec (0) = 1

sec (0)

sec (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} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} 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} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

= 1

= - 1

The two sides are not equal

\Rightarrow False

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

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

\frac{tan ^{2} ( x ) + 1}{1 + cot ^{2} ( x )} = tan^{2} (x)

3 cos (3 x) = 3 cos (x) - 12 sin^{2} (x) cos (x)

4 sec (y) cos (y) = 4