What is the value of (1+tan^2(67.5))/(1-tan^2(67.5)) ?

The value of (1+tan^2(67.5))/(1-tan^2(67.5)) is -sqrt(2)

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(1+tan^2(67.5))/(1-tan^2(67.5))

Solution

\frac{1 + tan ^{2} ( 67. 5 ^{\circ} )}{1 - tan ^{2} ( 67. 5 ^{\circ} )}

Solution

- 2

Decimal Notation

- 1.41421 \dots

Solution steps

\frac{1 + tan ^{2} ( 67. 5 ^{\circ} )}{1 - tan ^{2} ( 67. 5 ^{\circ} )}

Rewrite using trig identities:

tan^{2} (67. 5^{\circ}) = \frac{1 - cos ( 13 5 ^{\circ} )}{1 + cos ( 13 5 ^{\circ} )}

tan^{2} (67. 5^{\circ})

Use the following identity:

tan^{2} (θ) = \frac{1 - cos ( 2 θ )}{1 + cos ( 2 θ )}

Use the following identity

tan (θ) = \frac{sin ( θ )}{cos ( θ )}

Square both sides

tan^{2} (θ) = \frac{sin ^{2} ( θ )}{cos ^{2} ( θ )}

Rewrite using trig identities:

sin^{2} (θ) = \frac{1 - cos ( 2 θ )}{2}

Use the Double Angle identity

cos (2 θ) = 1 - 2 sin^{2} (θ)

Switch sides

2 sin^{2} (θ) - 1 = - cos (2 θ)

Add

1

to both sides

2 sin^{2} (θ) = 1 - cos (2 θ)

Divide both sides by

2

sin^{2} (θ) = \frac{1 - cos ( 2 θ )}{2}

Rewrite using trig identities:

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

Use the Double Angle identity

cos (2 θ) = 2 cos^{2} (θ) - 1

Switch sides

2 cos^{2} (θ) - 1 = cos (2 θ)

Add

1

to both sides

2 sin^{2} (θ) = 1 + cos (2 θ)

Divide both sides by

2

cos^{2} (θ) = \frac{1 + cos ( 2 θ )}{2}

tan^{2} (θ) = \frac{\frac{1 - c o s ( 2 θ )}{2}}{\frac{1 + c o s ( 2 θ )}{2}}

Simplify

tan^{2} (θ) = \frac{1 - cos ( 2 θ )}{1 + cos ( 2 θ )}

= \frac{1 - cos ( 2 \cdot 67. 5 ^{\circ} )}{1 + cos ( 2 \cdot 67. 5 ^{\circ} )}

Simplify

= \frac{1 - cos ( 13 5 ^{\circ} )}{1 + cos ( 13 5 ^{\circ} )}

= \frac{1 + \frac{1 - c o s ( 13 5 ^{\circ} )}{1 + c o s ( 13 5 ^{\circ} )}}{1 - \frac{1 - c o s ( 13 5 ^{\circ} )}{1 + c o s ( 13 5 ^{\circ} )}}

Simplify

= \frac{1}{cos ( 13 5 ^{\circ} )}

Use the following trivial identity:

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

cos (13 5^{\circ})

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} 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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= - \frac{2}{2}

= \frac{1}{- \frac{2}{2}}

Simplify

\frac{1}{- \frac{2}{2}} : - 2

\frac{1}{- \frac{2}{2}}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{1}{\frac{2}{2}}

Apply the fraction rule:

\frac{1}{\frac{b}{c}} = \frac{c}{b}

\frac{1}{\frac{2}{2}} = \frac{2}{2}

= - \frac{2}{2}

Apply radical rule:

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

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

Apply exponent rule:

\frac{x ^{a}}{x ^{b}} = x^{a - b}

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

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

Subtract the numbers:

1 - \frac{1}{2} = \frac{1}{2}

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

Apply radical rule:

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

= - 2

= - 2

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Frequently Asked Questions (FAQ)

What is the value of (1+tan^2(67.5))/(1-tan^2(67.5)) ?
The value of (1+tan^2(67.5))/(1-tan^2(67.5)) is -sqrt(2)