What is the value of tan^2(45-15) ?

The value of tan^2(45-15) is 1/3

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tan^2(45-15)

tan^{2} (4 5^{\circ} - 1 5^{\circ})

\frac{1}{3}

Decimal Notation

0.33333 \dots

Solution steps

tan^{2} (4 5^{\circ} - 1 5^{\circ})

Simplify

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

Use the following trivial identity:

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

tan (3 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}

= \frac{3}{3}

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

Simplify

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

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

Apply exponent rule:

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

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

(3)^{2} : 3

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

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

Cancel the common factor:

3

= \frac{1}{3}

= \frac{1}{3}

arctan (\frac{3}{- 2})

tan (8 6^{\circ})

sec (5 6^{\circ})

sin (11 7^{\circ})

1 - sin (\frac{11 π}{6})