Question

please someone solve this
please answer if you know otherwise will be reported​

Answer 1

Given Question :-

Solve the following :

$\sf \: 3 \: tan\bigg[ \theta \: - \dfrac{\pi}{12} \bigg] = tan\bigg[\theta + \dfrac{\pi}{12} \bigg]$

$\sf \: where \: 0 < \theta < \dfrac{\pi}{12}$

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

Given Trigonometric equation is

$\rm :\longmapsto\:\sf \: 3 \: tan\bigg[ \theta \: - \dfrac{\pi}{12} \bigg] = tan\bigg[\theta + \dfrac{\pi}{12} \bigg]$

Let assume that,

$\red{\rm :\longmapsto\:\dfrac{\pi}{12} = x}$

So, given equation can be rewritten as

$\rm :\longmapsto\:3tan(\theta - x) = tan(\theta + x)$

$\rm :\longmapsto\:3\dfrac{sin(\theta - x)}{cos(\theta - x)} = \dfrac{sin(\theta + x)}{cos(\theta + x)}$

$\rm :\longmapsto\:\dfrac{sin(\theta + x)cos(\theta - x)}{cos(\theta + x)sin(\theta - x)} = 3$

can be rewritten as

$\rm :\longmapsto\:\dfrac{sin(\theta + x)cos(\theta - x)}{cos(\theta + x)sin(\theta - x)} = \dfrac{3}{1}$

On applying Componendo and Dividendo, we get

$\rm :\longmapsto\:\dfrac{sin(\theta + x)cos(\theta - x) + cos(\theta + x)sin(\theta - x)}{sin(\theta + x)cos(\theta - x) - cos(\theta + x)sin(\theta - x)} = \dfrac{3 + 1}{3 - 1}$

We know,

$\boxed{ \bf{ \: sin(x + y) = sinxcosy + sinycosx}} \\ \\ \boxed{ \bf{ \: sin(x - y) = sinxcosy - sinycosx}}$

So, using this

$\rm :\longmapsto\:\dfrac{sin(\theta + x + \theta - x)}{sin(\theta + x - \theta + x)} = \dfrac{4}{2}$

$\rm :\longmapsto\:\dfrac{sin2\theta }{sin2x} = 2$

$\rm :\longmapsto\:sin2\theta = 2sin2x$

On substituting the value of x, we get

$\rm :\longmapsto\:sin2\theta = 2 \: sin\bigg[2 \times \dfrac{\pi}{12} \bigg]$

$\rm :\longmapsto\:sin2\theta = 2 \: sin\bigg[ \dfrac{\pi}{6} \bigg]$

$\rm :\longmapsto\:sin2\theta = 2 \: \times \dfrac{1}{2}$

$\rm :\longmapsto\:sin2\theta = 1$

As it is given that

$\boxed{ \bf{ \: 0 < \theta < \dfrac{\pi}{2}}}$

Therefore,

$\rm :\longmapsto\:sin2\theta = sin\dfrac{\pi}{2}$

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

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

Additional Information :-

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

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

$\boxed{ \bf{ \: tan(x + y) = \frac{tanx + tany}{1 - tanx \: tany}}}$

$\boxed{ \bf{ \: tan(x - y) = \frac{tanx - tany}{1 + tanx \: tany}}}$

$\boxed{ \bf{ \: sinx + siny = 2sin\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}$

$\boxed{ \bf{ \: sinx - siny = 2cos\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}$

$\boxed{ \bf{ \: cosx + cosy = 2cos\bigg[\dfrac{x + y}{2} \bigg]cos\bigg[\dfrac{x - y}{2} \bigg]}}$

$\boxed{ \bf{ \: cosx - cosy = - 2sin\bigg[\dfrac{x + y}{2} \bigg]sin\bigg[\dfrac{x - y}{2} \bigg]}}$