Is (1-sin(0))/(cos(0))=sec(0)-tan(0) ?

The answer to whether (1-sin(0))/(cos(0))=sec(0)-tan(0) is True

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prove (1-sin(0))/(cos(0))=sec(0)-tan(0)

prove

\frac{1 - sin ( 0 )}{cos ( 0 )} = sec (0) - tan (0)

True

Solution steps

\frac{1 - sin ( 0 )}{cos ( 0 )} = sec (0) - tan (0)

Manipulating left side

\frac{1 - sin ( 0 )}{cos ( 0 )}

Simplify

\frac{1 - sin ( 0 )}{cos ( 0 )} : 1

\frac{1 - sin ( 0 )}{cos ( 0 )}

1 - sin (0) = 1

1 - sin (0)

Use the following trivial identity:

sin (0) = 0

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 - 0

Subtract the numbers:

1 - 0 = 1

= 1

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

cos (0) = 1

cos (0)

Use the following trivial identity:

cos (0) = 1

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}

Apply rule

\frac{a}{1} = a

= 1

= 1

Manipulating right side

sec (0) - tan (0)

Simplify

sec (0) - tan (0) : 1

sec (0) - tan (0)

Use the following trivial identity:

sec (0) = 1

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 - tan (0)

Use the following trivial identity:

tan (0) = 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}

= 1 - 0

Simplify

= 1

= 1

We showed that the two sides could take the same form

\Rightarrow True

Is (1-sin(0))/(cos(0))=sec(0)-tan(0) ?
The answer to whether (1-sin(0))/(cos(0))=sec(0)-tan(0) is True