What is the value of tan^2(60)+2tan^2(45) ?

The value of tan^2(60)+2tan^2(45) is 5

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tan^2(60)+2tan^2(45)

tan^{2} (6 0^{\circ}) + 2 tan^{2} (4 5^{\circ})

5

Solution steps

tan^{2} (6 0^{\circ}) + 2 tan^{2} (4 5^{\circ})

Use the following trivial identity:

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

tan (6 0^{\circ})

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}

= 3

Use the following trivial identity:

tan (4 5^{\circ}) = 1

tan (4 5^{\circ})

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}

= 1

= (3)^{2} + 2 \cdot 1^{2}

Simplify

(3)^{2} + 2 \cdot 1^{2} : 5

(3)^{2} + 2 \cdot 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= (3)^{2} + 2 \cdot 1

(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

2 \cdot 1 = 2

2 \cdot 1

Multiply the numbers:

2 \cdot 1 = 2

= 2

= 3 + 2

Add the numbers:

3 + 2 = 5

= 5

= 5

What is the value of tan^2(60)+2tan^2(45) ?
The value of tan^2(60)+2tan^2(45) is 5