Question

evaluate : tan 13π\12

Answer 1

heya friend !!!

it's too easy ....

13π/12 =(π+π/12) we can write .

so , tan (13π/12)=tan(π+π/12)

=)tanπ/12. 【 tanπ/12 is in third quadrant so it is taken as in positive 】

=)tan(180°/12)

=)tan15*

tan15°=2-√3, -2-√3 Ans

=)(√3-1/√3+1) Ans .

hope it help you.☺

@rajukumar☺☺

Answer 2

The value of $\tan \dfrac{13\pi}{12}$ = 2 - $\sqrt{3}$

Step-by-step explanation:

We have,

$\tan \dfrac{13\pi}{12}$

To find, the value of $\tan \dfrac{13\pi}{12}$ = ?

∴ $\tan \dfrac{13\pi}{12}$

= $\tan (\pi+\dfrac{\pi}{12})$

We know that,

$\tan (\pi+A)= \tan A$ (since, 3rd co-ordinate tan are + ve)

$=\tan \dfrac{\pi}{12}$

=$\tan 15$

= $\dfrac{\sqrt{3}-1}{\sqrt{3}+1}$

Rationalising numerator and denominator, we get

= $\dfrac{\sqrt{3}-1}{\sqrt{3}+1}\times \dfrac{\sqrt{3}-1}{\sqrt{3}+1}$

= $\dfrac{(\sqrt{3}-1)^2}{\sqrt{3}^2-1^2}$

Using the algebraic identity,

$a^{2}-b^{2}=(a+b)(a-b)$

=$\dfrac{3+1-2\sqrt{3}}{3-1}$

Using the algebraic identity,

$(a-b)^{2}=a^{2}+b^{2} -2ab$

= 2 - $\sqrt{3}$

∴ The value of $\tan \dfrac{13\pi}{12}$ = 2 - $\sqrt{3}$