Is sec(30)-sin(30)tan(30)=cos(30) ?

The answer to whether sec(30)-sin(30)tan(30)=cos(30) is True

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Popular Trigonometry >

prove sec(30)-sin(30)tan(30)=cos(30)

Solution

prove

sec (3 0^{\circ}) - sin (3 0^{\circ}) tan (3 0^{\circ}) = cos (3 0^{\circ})

Solution

True

Solution steps

sec (3 0^{\circ}) - sin (3 0^{\circ}) tan (3 0^{\circ}) = cos (3 0^{\circ})

Manipulating left side

sec (3 0^{\circ}) - sin (3 0^{\circ}) tan (3 0^{\circ})

Simplify

sec (3 0^{\circ}) - sin (3 0^{\circ}) tan (3 0^{\circ}) : \frac{3}{2}

sec (3 0^{\circ}) - sin (3 0^{\circ}) tan (3 0^{\circ})

sec (3 0^{\circ}) = \frac{2 3}{3}

sec (3 0^{\circ})

Use the following trivial identity:

sec (3 0^{\circ}) = \frac{2 3}{3}

sec (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sec (x) 1 \frac{2 3}{3} 22 Undefined - 2 - 2 - \frac{2 3}{3} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sec (x) - 1 - \frac{2 3}{3} - 2 - 2 Undefined 22 \frac{2 3}{3}

= \frac{2 3}{3}

sin (3 0^{\circ}) tan (3 0^{\circ}) = \frac{3}{6}

sin (3 0^{\circ}) tan (3 0^{\circ})

Simplify

sin (3 0^{\circ}) : \frac{1}{2}

sin (3 0^{\circ})

Use the following trivial identity:

sin (3 0^{\circ}) = \frac{1}{2}

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

= \frac{1}{2} tan (3 0^{\circ})

Simplify

tan (3 0^{\circ}) : \frac{3}{3}

tan (3 0^{\circ})

Use the following trivial identity:

tan (3 0^{\circ}) = \frac{3}{3}

tan (x)

periodicity table with

18 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

= \frac{3}{3}

= \frac{1}{2} \cdot \frac{3}{3}

Multiply fractions:

\frac{a}{b} \cdot \frac{c}{d} = \frac{a \cdot c}{b \cdot d}

= \frac{3}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{3}{6}

= \frac{2 3}{3} - \frac{3}{6}

Simplify

\frac{2 3}{3} - \frac{3}{6}

Least Common Multiplier of

3, 6 : 6

3, 6

Least Common Multiplier (LCM)

Prime factorization of

3 : 3

3

3

is a prime number, therefore no factorization is possible

= 3

Prime factorization of

6 : 2 \cdot 3

6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 3

Multiply each factor the greatest number of times it occurs in either

3

6

= 3 \cdot 2

Multiply the numbers:

3 \cdot 2 = 6

= 6

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

6

For

\frac{2 3}{3} :

multiply the denominator and numerator by

2

\frac{2 3}{3} = \frac{2 3 \cdot 2}{3 \cdot 2} = \frac{4 3}{6}

= \frac{4 3}{6} - \frac{3}{6}

Since the denominators are equal, combine the fractions:

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

= \frac{4 3 - 3}{6}

Add similar elements:

43 - 3 = 33

= \frac{3 3}{6}

Cancel the common factor:

3

= \frac{3}{2}

= \frac{3}{2}

= \frac{3}{2}

Manipulating right side

cos (3 0^{\circ})

Simplify

= \frac{3}{2}

We showed that the two sides could take the same form

\Rightarrow True

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Frequently Asked Questions (FAQ)

Is sec(30)-sin(30)tan(30)=cos(30) ?
The answer to whether sec(30)-sin(30)tan(30)=cos(30) is True