What is the value of tan^2(240)+sec^2(225) ?

The value of tan^2(240)+sec^2(225) is 5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tan^2(240)+sec^2(225)

tan^{2} (24 0^{\circ}) + sec^{2} (22 5^{\circ})

5

Solution steps

tan^{2} (24 0^{\circ}) + sec^{2} (22 5^{\circ})

tan (24 0^{\circ}) = tan (6 0^{\circ})

tan (24 0^{\circ})

Rewrite

24 0^{\circ}

18 0^{\circ} + 6 0^{\circ}

= tan (18 0^{\circ} + 6 0^{\circ})

Apply the periodicity of

tan

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

tan (18 0^{\circ} + 6 0^{\circ}) = tan (6 0^{\circ})

= tan (6 0^{\circ})

= tan^{2} (6 0^{\circ}) + sec^{2} (22 5^{\circ})

Rewrite using trig identities:

sec^{2} (22 5^{\circ}) = tan^{2} (4 5^{\circ}) + 1

sec^{2} (22 5^{\circ})

Use the Pythagorean identity:

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

= tan^{2} (22 5^{\circ}) + 1

tan (22 5^{\circ}) = tan (4 5^{\circ})

tan (22 5^{\circ})

Rewrite

22 5^{\circ}

18 0^{\circ} + 4 5^{\circ}

= tan (18 0^{\circ} + 4 5^{\circ})

Apply the periodicity of

tan

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

tan (18 0^{\circ} + 4 5^{\circ}) = tan (4 5^{\circ})

= tan (4 5^{\circ})

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

= tan^{2} (6 0^{\circ}) + tan^{2} (4 5^{\circ}) + 1

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} + 1^{2} + 1

Simplify

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

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

Apply rule

1^{a} = 1

1^{2} = 1

= (3)^{2} + 1 + 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

= 3 + 1 + 1

Add the numbers:

3 + 1 + 1 = 5

= 5

= 5

70 cos (3 0^{\circ})

1 + 2 cos (\frac{π}{3})

arctan (56)

sin (\frac{π}{4}) \cdot sin (\frac{π}{4})

ln (2 + tan (\frac{π}{32}))

What is the value of tan^2(240)+sec^2(225) ?
The value of tan^2(240)+sec^2(225) is 5