# If 2A=tan^-1 (4/3), find the value of tan(A-arccos-1/2)?

Jun 21, 2017

$\text{ The Reqd. Value=} 8 + 5 \sqrt{3.}$

#### Explanation:

In this Solution, we assume that, as functions ${\tan}^{-} 1 = a r c \tan .$

Let us first find, $a r c \cos \left(- \frac{1}{2}\right) .$

Recall the following Definition of $a r c \cos$ function :

$a r c \cos x = \theta , | x | \le 1 \iff \cos \theta = x , \theta \in \left[0 , \pi\right] .$

We have, cos(2pi/3)=cos (pi-pi/3)=-cos(pi/3)=-1/2, &, (2pi/3) in [0,pi].

$\Rightarrow a r c \cos \left(- \frac{1}{2}\right) = 2 \frac{\pi}{3.} \ldots \ldots \ldots . \text{[Defn.]}$

Therefore, the Reqd. Value=$\tan \left(A - a r c \cos \left(- \frac{1}{2}\right)\right) ,$

$= \tan \left(A - 2 \frac{\pi}{3}\right) ,$

$\therefore \text{ the Reqd. Value=} \frac{\tan A - \tan \left(2 \frac{\pi}{3}\right)}{1 + \tan A \tan \left(2 \frac{\pi}{3}\right)} \ldots \ldots . . \left(\ast\right) .$

Here, $\tan \left(2 \frac{\pi}{3}\right) = \tan \left(\pi - \frac{\pi}{3}\right) = - \tan \left(\frac{\pi}{3}\right) = - \sqrt{3.} \ldots . \left({\ast}_{1}\right) .$

Given that, 2A=tan^-1(4/3) rArr tan2A=4/3, &, 2A in (-pi/2,pi/2).

But, $\tan 2 A > 0 , 2 A \in \left(0 , \frac{\pi}{2}\right) .$

$\because , \tan 2 A = \frac{2 \tan A}{1 - {\tan}^{2} A} = \frac{2 t}{1 - {t}^{2}} , t = \tan A ,$

$\therefore \frac{2 t}{1 - {t}^{2}} = \frac{4}{3} \therefore 2 - 2 {t}^{2} = 3 t \therefore 2 {t}^{2} + 3 t - 2 = 0.$

$\therefore \left(t + 2\right) \left(2 t - 1\right) = 0 \therefore t = - 2 , \mathmr{and} , t = \frac{1}{2.}$

Since, $2 A \in \left(0 , \frac{\pi}{2}\right) , t = \tan A > 0 \Rightarrow t = \frac{1}{2.} \ldots \left({\ast}_{2}\right) .$

Utilising $\left({\ast}_{1}\right) \mathmr{and} \left({\ast}_{2}\right) \text{ in } \left(\ast\right) ,$ we have,

$\text{ The Reqd. Value=} \frac{\frac{1}{2} - \left(- \sqrt{3}\right)}{1 + \left(\frac{1}{2}\right) \left(- \sqrt{3}\right)} = \frac{1 + 2 \sqrt{3}}{2 - \sqrt{3}}$

$= \left(1 + 2 \sqrt{3}\right) \left(2 + \sqrt{3}\right) = 8 + 5 \sqrt{3.}$

Enjoy Maths.!