Question

If 6cos²Ø-2cos2Ø-3=0,then tan²3Ø= ?

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Answer 1

We know,

$\cos(2\theta)=2\cos^2\theta-1$

So,

$6\cos^2\theta-2\cos(2\theta)-3=0\\\\\\\implies\ 6\cos^2\theta-2[2\cos^2\theta-1]-3=0\\\\\\\implies\ 6\cos^2\theta-4\cos^2\theta+2-3=0\\\\\\\implies\ 2\cos^2\theta-1=0\\\\\\\implies\ \cos(2\theta)=\cos\left(\dfrac{\pi}{2}\right)\\\\\\\implies\ 2\theta=2n\pi\pm\dfrac{\pi}{2}\\\\\\\implies\ 3\theta=3n\pi\pm\dfrac{3\pi}{4}\ \equiv\ n\pi\pm\dfrac{3\pi}{4}\\\\\\\implies\ \tan(3\theta)=\pm\tan\left(\dfrac{3\pi}{4}\right)\ =\ \pm1\\\\\\\implies\ \tan^2(3\theta)=\mathbf{1}$

Hence 1 is the answer.

In this solution, there may be a doubt on how  $3n\pi\pm\dfrac{3\pi}{4}\ \equiv\ n\pi\pm\dfrac{3\pi}{4}.$

It is true that we reach the same position for every  $2n\pi$  rotations, either clockwise or anti-clockwise, but anti-clockwise is taken here. And we know the initial position is at the x - axis.

That's why  $\tan(2n\pi+x),$  for example, is simply taken as  $\tan x.$

Here this means  $2n\pi\equiv0.$

This implies  $3n\pi=2n\pi+n\pi\equiv0+n\pi=n\pi.$