Question

If cot(A+B).cot(A-B)=1 then the value of cot(2A/ 3)

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{cot(A+B)\cdot\,cot(A-B)=1}$

$\sf{\implies\,tan(A+B)\cdot\,tan(A-B)=1}$

$\sf{\implies\,\dfrac{tan(A)+tan(B)}{1-tan(A)\,tan(B)}\cdot\,\dfrac{tan(A)-tan(B)}{1+tan(A)\,tan(B)}=1}$

$\sf{\implies\,\dfrac{tan^2(A)-tan^2(B)}{1-tan^2(A)\,tan^2(B)}=1}$

$\sf{\implies\,tan^2(A)-tan^2(B)=1-tan^2(A)\,tan^2(B)}$

$\sf{\implies\,tan^2(A)+tan^2(A)\,tan^2(B)-tan^2(B)-1=0}$

$\sf{\implies\,tan^2(A)(1+tan^2(B))-1(1+tan^2(B))=0}$

$\sf{\implies\,(tan^2(A)-1)(tan^2(B)+1)=0}$

$\sf{\implies\,tan^2(A)=1\,\,\,\,\,\,or\,\,\,\,\,\,tan^2(B)=-1}$

tan²(x) can not be negative, so,

$\sf{\implies\,tan^2(A)=1}$

$\sf{\implies\,A=\dfrac{\pi}{4}}$

Now,

$\sf{cot\bigg(\dfrac{2}{3}\cdot\dfrac{\pi}{4}\bigg)}$

$\sf{=cot\bigg(\dfrac{1}{3}\cdot\dfrac{\pi}{2}\bigg)}$

$\sf{=cot\bigg(\dfrac{\pi}{6}\bigg)}$

$\sf{=\sqrt{3}}$

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:cot(A + B)cot(A - B) = 1$

We know,

$\boxed{ \tt{ \: cotx = \frac{cosx}{sinx} \: }}$

So, using this identity, we get

$\rm :\longmapsto\:\dfrac{cos(A + B)}{sin(A + B)} \times \dfrac{cos(A - B)}{sin(A - B)} = 1$

$\rm :\longmapsto\:cos(A + B)cos(A - B) = sin(A + B)sin(A - B)$

$\rm :\longmapsto\:cos(A + B)cos(A - B) - sin(A + B)sin(A - B) = 0$

We know,

$\boxed{ \tt{ \: cosxcosy - sinxsiny = cos(x + y) \: }}$

So, using this, we get

$\rm :\longmapsto\:cos(A + B + A - B) = 0$

$\rm :\longmapsto\:cos(2A) = 0$

$\rm :\longmapsto\:cos(2A) =cos\dfrac{\pi}{2}$

$\rm :\longmapsto\:2A = \dfrac{\pi}{2}$

$\bf\implies \:A = \dfrac{\pi}{4}$

Now, Consider

$\rm :\longmapsto\:cot\dfrac{2A}{3}$

$\rm \:  =  \:cot\bigg[\dfrac{2}{3} \times \dfrac{\pi}{4} \bigg]$

$\rm \:  =  \:cot\dfrac{\pi}{6}$

$\rm \:  =  \: \sqrt{3}$

More to know :-

$\purple{\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf T-eq & \bf Solution \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf sinx = 0 & \sf x = n\pi \: \forall \: n \in \: Z\\ \\ \sf cosx = 0 & \sf x = (2n + 1)\dfrac{\pi}{2}\: \forall \: n \in \: Z\\ \\ \sf tanx = 0 & \sf x = n\pi\: \forall \: n \in \: Z\\ \\ \sf sinx = siny & \sf x = n\pi + {( - 1)}^{n}y \: \forall \: n \in \: Z\\ \\ \sf cosx = cosy & \sf x = 2n\pi \pm \: y\: \forall \: n \in \: Z\\ \\ \sf tanx = tany & \sf x = n\pi + y \: \forall \: n \in \: Z\end{array}} \\ \end{gathered}\end{gathered}}$