$(sec a + 1) = tan a (2 + \sqrt 3)$ $(sec a+1)=tana(2+√3)$ 0<=a<=2π??

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$(sec a + 1) = tan a (2 + \sqrt 3)$ $(sec a+1)=tana(2+√3)$ 0<=a<=2π??

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Answer 1 · 2018-02-02T07:19:46

$(sec a + 1) = tan a (2 + \sqrt 3)$ $(sec a+1)=tana(2+√3)$

$\Rightarrow (sec a + 1) \times cos a = cos a \times tan a (2 + \sqrt 3)$ $=>(seca+1)xxcosa=cosaxxtana(2+√3)$

$\Rightarrow (cos a + 1) = sin a (2 + \sqrt 3)$ $=>(cosa+1)=sina(2+√3)$

$\Rightarrow (2 + \sqrt 3) 2 sin (\frac{a}{2}) cos (\frac{a}{2}) = 2 {cos}^{2} (\frac{a}{2})$ $=>(2+√3)2sin(a/2)cos(a/2)=2cos^2(a/2)$

$\Rightarrow [(2 + \sqrt 3) sin (\frac{a}{2}) - cos (\frac{a}{2})] 2 cos (\frac{a}{2}) = 0$ $=>[(2+√3)sin(a/2)-cos(a/2)]2cos(a/2)=0$

So $cos (\frac{a}{2}) = 0 = cos (\frac{π}{2})$ $cos (a/2)=0=cos(pi/2)$

$\Rightarrow \frac{a}{2} = \frac{π}{2}$ $=>a/2=pi/2$

$\Rightarrow a = π$ $=>a=pi$

Again
$[(2 + \sqrt 3) sin (\frac{a}{2}) - cos (\frac{a}{2})] = 0$ $[(2+√3)sin(a/2)-cos(a/2)]=0$

$\Rightarrow (2 + \sqrt 3) sin (\frac{a}{2}) = cos (\frac{a}{2})]$ $=>(2+√3)sin(a/2)=cos(a/2)]$

$\Rightarrow \frac{sin (\frac{a}{2})}{cos (\frac{a}{2})} = \frac{1}{2 + \sqrt 3}$ $=>sin(a/2)/cos(a/2)=1/(2+√3)$

$\Rightarrow tan (\frac{a}{2}) = 2 - \sqrt{3} = tan (\frac{π}{12})$ $=>tan(a/2)=2-sqrt3=tan(pi/12)$ $see note below$ $color(red)"see note below"$

$\Rightarrow \frac{a}{2} = \frac{π}{12}$ $=>a/2=pi/12$

$\Rightarrow a = \frac{π}{6}$ $=>a=pi/6$

Alternative method

$(sec a + 1) = tan a (2 + \sqrt 3)$ $(sec a+1)=tana(2+√3)$

$\Rightarrow (sec a + 1) - tan a (2 + \sqrt 3) = 0$ $=>(sec a+1)-tana(2+√3)=0$

$\Rightarrow {(\sqrt{sec a + 1})}^{2} - (\sqrt{{sec}^{2} a - 1}) (2 + \sqrt 3) = 0$ $=>(sqrt(sec a+1))^2-(sqrt(sec^2a-1))(2+√3)=0$

$\Rightarrow (\sqrt{sec a + 1}) [\sqrt{sec a + 1} - (\sqrt{sec a - 1}) (2 + \sqrt 3)] = 0$ $=>(sqrt(sec a+1))[sqrt(seca+1)-(sqrt(seca-1))(2+√3)]=0$

So $\sqrt{sec a + 1} = 0$ $sqrt(sec a+1)=0$

$\Rightarrow sec a = - 1$ $=>sec a=-1$

$\Rightarrow cos a = - 1 = cos π$ $=>cosa=-1=cospi$

When $0 \leq a \leq 2 π$ $0<= a <= 2pi$
$a = π$ $a=pi$

Again

$\sqrt{sec a + 1} - (\sqrt{sec a - 1}) (2 + \sqrt 3) = 0$ $sqrt(seca+1)-(sqrt(seca-1))(2+√3)=0$

$\Rightarrow \sqrt{\frac{sec a + 1}{sec a - 1}} = (2 + \sqrt 3)$ $=>sqrt((seca+1)/(seca-1))=(2+√3)$

$\Rightarrow \sqrt{\frac{1 + cos a}{1 - cos a}} = (2 + \sqrt 3)$ $=>sqrt((1+cosa)/(1-cosa))=(2+√3)$

$\Rightarrow \sqrt{\frac{2 {cos}^{2} (\frac{a}{2})}{2 {sin}^{2} (\frac{a}{2})}} = (2 + \sqrt 3)$ $=>sqrt((2cos^2(a/2))/(2sin^2(a/2)))=(2+√3)$

$\Rightarrow tan (\frac{a}{2}) = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3} = tan (\frac{π}{12})$ $=>tan(a/2)=1/(2+sqrt3)=2-sqrt3=tan(pi/12)$

$\Rightarrow \frac{a}{2} = \frac{π}{12}$ $=>a/2=pi/12$

$\Rightarrow a = \frac{π}{6}$ $=>a=pi/6$

Again
$\Rightarrow tan (\frac{a}{2}) = tan (\frac{π}{12}) = tan (π + \frac{π}{12}) = tan (\frac{13 π}{12})$ $=>tan(a/2)=tan(pi/12)=tan(pi+pi/12)=tan((13pi)/12)$

then $a = \frac{13 π}{6}$ $a=(13pi)/6$ being $> 2 π$ $>2pi$ the solution neglected.

Note

$tan (\frac{π}{12}) = tan (\frac{π}{4} - \frac{π}{6})$ $tan(pi/12)=tan(pi/4-pi/6)$

$= \frac{tan (\frac{π}{4}) - tan (\frac{π}{6})}{1 - tan (\frac{π}{4}) \cdot tan (\frac{π}{6})}$ $=(tan(pi/4)-tan(pi/6))/(1-tan(pi/4)*tan(pi/6))$

$= \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}}$ $=(1-1/sqrt3)/(1+1*1/sqrt3)$

$= \frac{\sqrt{3} - 1}{\sqrt{3} + 1}$ $=(sqrt3-1)/(sqrt3+1)$

$= \frac{{(\sqrt{3} - 1)}^{2}}{{(\sqrt{3})}^{2} - 1^{2}}$ $=(sqrt3-1)^2/((sqrt3)^2-1^2)$

$= \frac{4 - 2 \sqrt{3}}{3 - 1}$ $=(4-2sqrt3)/(3-1)$

$= \frac{4 - 2 \sqrt{3}}{2}$ $=(4-2sqrt3)/2$

$= 2 - \sqrt{3}$ $=2-sqrt3$

Answer 2 · 2018-02-02T11:17:39

$a \in {\frac{π}{6}, π} \subset [0, 2 π]$ $a in {pi/6,pi} sub [0,2pi]$ .

Explanation:

We have, ${sec}^{2} a - 1 = tan a$ $sec^2a-1=tana$ .

$:. (seca+1)(seca-1)=tan^2a$ .

Sub.ing the given value, $(seca+1)=(2+sqrt3)tana...(1)$ , we get,

$(2+sqrt3)tana(seca-1)=tan^2a, or,$

$tan^2a-(2+sqrt3)tana(seca-1)=0$ .

$:. tana{tana-(2+sqrt3)(seca-1)}=0$ .

$:. tana=0, or, tana=(2+sqrt3)(seca-1)$ .

Case 1 : $tana=0$ .

$tana=0" with, "a in [0,2pi] rArr a=0,pi,2pi$ .

But, $a=0 and 2pi$ do not satisfy the given eqn.

$:. a=pi$ .

Case 2 : $tana=(2+sqrt3)(seca-1)$ .

$tana=(2+sqrt3)(seca-1) rArr seca-1=tana/(2+sqrt3)$ .

$:. seca-1=(2-sqrt3)tana............(2)$ .

$(1)+(2) rArr 2seca=4tana, i.e., sina=1/2$ .

$sina=1/2," read with "a in [0,2pi] rArr a=pi/6, (pi-pi/6)=5pi/6$

Here again, $a=5pi/6$ is extraneous.

$:. a=pi/6$ .

Altogether, $a in {pi/6,pi} sub [0,2pi]$ .

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$(sec a + 1) = tan a (2 + \sqrt 3)$ $(sec a+1)=tana(2+√3)$ 0<=a<=2π??

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