What is the value of cube root of 5cot^2(pi/6)+3sec^2(pi/3) ?

The value of cube root of 5cot^2(pi/6)+3sec^2(pi/3) is 3

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cube root of 5cot^2(pi/6)+3sec^2(pi/3)

3

Solution steps

Use the following trivial identity:

cot (\frac{π}{6}) = 3

cot (\frac{π}{6})

cot (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} cot (x) \mp \infty 31 \frac{3}{3} 0 - \frac{3}{3} - 1 - 3

= 3

Use the following trivial identity:

sec (\frac{π}{3}) = 2

sec (\frac{π}{3})

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}

= 2

Simplify

5 (3)^{2} = 15

5 (3)^{2}

(3)^{2} = 3

(3)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 3

= 5 \cdot 3

Multiply the numbers:

5 \cdot 3 = 15

= 15

3 \cdot 2^{2} = 12

3 \cdot 2^{2}

2^{2} = 4

= 3 \cdot 4

Multiply the numbers:

3 \cdot 4 = 12

= 12

Add the numbers:

15 + 12 = 27

Factor the number:

27 = 3^{3}

Apply radical rule:

= 3

= 3

csc (\frac{5}{4} π)

\frac{5 sin ( 14 0 ^{\circ} )}{7}

- sin (\frac{11 π}{6})

tan (\frac{7}{4} π)

\frac{π}{2} - cos (\frac{π}{2})

What is the value of cube root of 5cot^2(pi/6)+3sec^2(pi/3) ?
The value of cube root of 5cot^2(pi/6)+3sec^2(pi/3) is 3