Question

If $cos2\theta +9sin2\theta -6sin\theta +54cos\theta =1$ then the value of $100ta{n}^2\theta +9tan\theta$ is equal to

Answer 1

$\cos 2\theta +9\sin 2\theta -6\sin \theta +54\cos \theta =1$

${\cos }^2\theta -{\sin }^2\theta +9\sin 2\theta -6\sin \theta +54\cos \theta =1$

Dividing by ${\cos }^2\theta$ , we get

$\Rightarrow 2{\tan }^2\theta -18\tan \theta +6\tan \theta \sec \theta -54\sec \theta =0$

$\Rightarrow (\tan \theta -9)(2\tan \theta +6\sec \theta )=0$

$\Rightarrow \tan \theta =9$ or $2\sin \theta +6=0\Rightarrow \sin \theta =-3$(not possible)

Hence, $\tan \theta =9$

Consider, $100ta{n}^2\theta +9tan\theta$

$100\left(81\right)+81=8181$