What is the value of (2tan(105))/(1-tan^2(105)) ?

The value of (2tan(105))/(1-tan^2(105)) is (sqrt(3))/3

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(2tan(105))/(1-tan^2(105))

\frac{2 tan ( 10 5 ^{\circ} )}{1 - tan ^{2} ( 10 5 ^{\circ} )}

\frac{3}{3}

Decimal Notation

0.57735 \dots

Solution steps

\frac{2 tan ( 10 5 ^{\circ} )}{1 - tan ^{2} ( 10 5 ^{\circ} )}

Rewrite using trig identities:

tan (3 0^{\circ})

\frac{2 tan ( 10 5 ^{\circ} )}{1 - tan ^{2} ( 10 5 ^{\circ} )}

Use the Double Angle identity:

\frac{2 tan ( x )}{1 - tan ^{2} ( x )} = tan (2 x)

= tan (2 \cdot 10 5^{\circ})

Simplify

= tan (21 0^{\circ})

tan (21 0^{\circ}) = tan (3 0^{\circ})

tan (21 0^{\circ})

Rewrite

21 0^{\circ}

18 0^{\circ} + 3 0^{\circ}

= tan (18 0^{\circ} + 3 0^{\circ})

Apply the periodicity of

tan

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

tan (18 0^{\circ} + 3 0^{\circ}) = tan (3 0^{\circ})

= tan (3 0^{\circ})

= tan (3 0^{\circ})

= tan (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}

- sin (2 \frac{π}{12}) \cdot 2

cos (5 \cdot \frac{π}{2})

sin (4 5^{\circ}) + sin (4 5^{\circ})

arcsin (\frac{1}{1.25})

arctan (sec (0))

What is the value of (2tan(105))/(1-tan^2(105)) ?
The value of (2tan(105))/(1-tan^2(105)) is (sqrt(3))/3