The answer to whether 1/2 sec(x)-1=0 is False

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prove 1/2 sec(x)-1=0

Solution

prove

\frac{1}{2} sec (x) - 1 = 0

False

Solution steps

\frac{1}{2} sec (x) - 1 = 0

Show that the two sides are not equal

For

\frac{1}{2} sec (x) - 1 = 0

plug in

x = 0

\frac{1}{2} sec (0) - 1 = - \frac{1}{2} (Decimal : - 0.5)

\frac{1}{2} sec (0) - 1

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

= \frac{1}{2} \cdot 1 - 1

Simplify

\frac{1}{2} \cdot 1 - 1 : - \frac{1}{2}

\frac{1}{2} \cdot 1 - 1

Multiply:

\frac{1}{2} \cdot 1 = \frac{1}{2}

= \frac{1}{2} - 1

Convert element to fraction:

1 = \frac{1 \cdot 2}{2}

= - \frac{1 \cdot 2}{2} + \frac{1}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{- 1 \cdot 2 + 1}{2}

- 1 \cdot 2 + 1 = - 1

- 1 \cdot 2 + 1

Multiply the numbers:

1 \cdot 2 = 2

= - 2 + 1

Add/Subtract the numbers:

- 2 + 1 = - 1

= - 1

= \frac{- 1}{2}

Apply the fraction rule:

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

= - \frac{1}{2}

= - \frac{1}{2}

The two sides are not equal

\Rightarrow False