Is arccos(x)= 1/(cos(x)) ?

The answer to whether arccos(x)= 1/(cos(x)) is False

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prove arccos(x)= 1/(cos(x))

prove

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

False

Solution steps

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

Show that the two sides are not equal

For

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

plug in

x = 0

arccos (0) = \frac{π}{2} (Decimal : Degrees : 1.57079 \dots 9 0^{\circ})

arccos (0)

Use the following trivial identity:

arccos (0) = \frac{π}{2}

arccos (0)

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2}

= \frac{π}{2}

\frac{1}{cos ( 0 )} = 1

\frac{1}{cos ( 0 )}

Use the following trivial identity:

cos (0) = 1

cos (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}

= 1

= \frac{1}{1}

Simplify

= 1

The two sides are not equal

\Rightarrow False

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

cos^{2} (x) (1 - sec^{2} (x)) = - sin^{2} (x)

sin^{2} (θ) - cos^{2} (θ) = 1

tan^{2} (θ) csc^{2} (θ) = sec^{2} (θ)

(1 + tan (x)) (1 - cot (x)) = tan (x) - cot (x)