What is the value of tan^2((7pi)/4) ?

The value of tan^2((7pi)/4) is 1

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tan^2((7pi)/4)

tan^{2} (\frac{7 π}{4})

1

Solution steps

tan^{2} (\frac{7 π}{4})

tan (\frac{7 π}{4}) = tan (\frac{3 π}{4})

tan (\frac{7 π}{4})

Rewrite

\frac{7 π}{4}

π + \frac{3 π}{4}

= tan (π + \frac{3 π}{4})

Apply the periodicity of

tan

tan (x + π) = tan (x)

tan (π + \frac{3 π}{4}) = tan (\frac{3 π}{4})

= tan (\frac{3 π}{4})

= tan^{2} (\frac{3 π}{4})

Rewrite using trig identities:

sec^{2} (\frac{3 π}{4}) - 1

tan^{2} (\frac{3 π}{4})

Use the Pythagorean identity:

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

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

= sec^{2} (\frac{3 π}{4}) - 1

= sec^{2} (\frac{3 π}{4}) - 1

Use the following trivial identity:

sec (\frac{3 π}{4}) = - 2

sec (\frac{3 π}{4})

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

= (- 2)^{2} - 1

Simplify

(- 2)^{2} - 1 : 1

(- 2)^{2} - 1

(- 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 - 1

Subtract the numbers:

2 - 1 = 1

= 1

= 1

cos (2) + 1 - 2

3 (\frac{2 π}{3}) + 2 cos (\frac{2 π}{3})

- 6 sin (\frac{2 π}{0.8} (0.5 - 0.2)) + 15

sin^{2} (12 0^{\circ}) - cos^{2} (12 0^{\circ})

\frac{9.8}{0.0272 \cdot cos ^{2} ( - 0.78 )}