Question

evaluate : tan 13π\12 answer this question too thanks to you all who have answered my questions.

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Answer 1

Answer:

Tan13π/12

Tan-195°

Tan-180+15

Tan-15

Tan-2-root 3 answer

Answer 2

Answer:

All trigonometric functions are periodic in nature. Hence they generally repeat after a certain interval.

In the general way, all trigonometric functions commonly remain periodic from (0, 2π).

Hence angles greater than 2π can be expressed in terms of angle A which is between 0 and 2π itself.

Now according to the question,

Tan ( 13π/12 ) is equivalent to another angle which lies between (0, 2π).

Expressing 13π/12 in the given interval we get:

$\implies \dfrac{13 \pi }{12} = \dfrac{ 24 \pi }{12} - \dfrac{11 \pi}{12}\\\\\\\implies \dfrac{13 \pi }{12} = 2 \pi - \dfrac{11 \pi}{12}\\\\\\\text{Therefore the angle expressed in between the interval is: } \\\\\\\implies \boxed{ \bf{ Tan( \:\dfrac{11\pi}{12})}}$

$\implies \dfrac{11\pi}{12} = \dfrac{11 \times 180^\circ}{12}\\\\\\\implies \boxed{ \bf{ Angle = 165^\circ}}$

⇒ Tan ( 165 )° = Tan ( 180 - 15 )° = Tan 15°

Now we know that,

⇒ Tan 15° = Tan ( 45° - 30° )

Applying Tan ( A - B ) formula we get:

$\implies \boxed{ Tan(A-B) = \dfrac{ Tan\:A - Tan\:B}{1 + Tan\:A.\:Tan\:B}}$

$\implies Tan (45 - 30 ) = \dfrac{ Tan\:45 - Tan\:30 }{ 1 + Tan\:45.\:Tan\:30}\\\\\\\implies Tan\:15^\circ = \dfrac{ 1 - \sqrt{3} }{ 1 + 1.\sqrt{3}}\\\\\\\implies \boxed{ \bf{ Tan\:15^\circ = \dfrac{ 1 - \sqrt{3} }{ 1 + \sqrt{3} }}}$

This is the required answer.