Is tan^2(x)-sec(x)=1 ?

The answer to whether tan^2(x)-sec(x)=1 is False

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prove tan^2(x)-sec(x)=1

prove

tan^{2} (x) - sec (x) = 1

False

Solution steps

tan^{2} (x) - sec (x) = 1

Show that the two sides are not equal

For

tan^{2} (x) - sec (x) = 1

plug in

x = 0

tan^{2} (0) - sec (0) = - 1

tan^{2} (0) - sec (0)

Use the following trivial identity:

tan (0) = 0

tan (0)

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= 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

= 0^{2} - 1

Simplify

= - 1

The two sides are not equal

\Rightarrow False

2 sin (π) = 5 sin (π)

tan (x) - \frac{1}{tan ( x )} = 2 sin^{2} (x) - 1

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

sin^{2} (x) \cdot sec (x) = tan^{2} (x)

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