What is the value of 1+tan^2(315) ?

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1+tan^2(315)

1 + tan^{2} (31 5^{\circ})

2

Solution steps

1 + tan^{2} (31 5^{\circ})

tan (31 5^{\circ}) = tan (13 5^{\circ})

tan (31 5^{\circ})

Rewrite

31 5^{\circ}

18 0^{\circ} + 13 5^{\circ}

= tan (18 0^{\circ} + 13 5^{\circ})

Apply the periodicity of

tan

tan (x + 18 0^{\circ}) = tan (x)

tan (18 0^{\circ} + 13 5^{\circ}) = tan (13 5^{\circ})

= tan (13 5^{\circ})

= 1 + tan^{2} (13 5^{\circ})

Rewrite using trig identities:

sec^{2} (13 5^{\circ})

tan^{2} (13 5^{\circ}) + 1

Use the Pythagorean identity:

tan^{2} (x) + 1 = sec^{2} (x)

= sec^{2} (13 5^{\circ})

= sec^{2} (13 5^{\circ})

Use the following trivial identity:

sec (13 5^{\circ}) = - 2

sec (13 5^{\circ})

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}

= - 2

= (- 2)^{2}

Simplify

(- 2)^{2} : 2

(- 2)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 2)^{2} = (2)^{2}

= (2)^{2}

Apply radical rule:

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

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

Apply exponent rule:

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

= 2^{\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

= 2

= 2

- cos (4 \frac{π}{2})

arctan (\frac{16}{32})

- 2 (3.27) - e^{- 0.54} sin (3.27)

2 \cdot cos (\frac{3 π}{2})

sec (3 0^{\circ}) - csc (2 4^{\circ})